# NEET Physics Two Dimensional Motion Maximum Height

Maximum Height

Category : NEET

Maximum Height Maximum height: It is the maximum height from the point of projection, a projectile can reach. So, by using ${{v}^{2}}={{u}^{2}}+2as$ $0={{(u\sin \theta )}^{2}}-2gH$ $H=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}$ (i) Maximum height can also be expressed as $H=\frac{u_{y}^{2}}{2g}$  (where ${{u}_{y}}$ is the vertical component of initial velocity). (ii) ${{H}_{\max }}=\frac{{{u}^{2}}}{2g}$  (when ${{\sin }^{2}}\,\theta = max = 1$ i.e., $\theta = 9{{0}^{o}}$) i.e., for maximum height body should be projected vertically upward. So it falls back to the point of projection after reaching the maximum height. (iii) For complementary angles of projection $\theta \,\,and\, 9{{0}^{o}}\theta$ Ratio of maximum height = $\frac{{{H}_{1}}}{{{H}_{2}}}=\frac{{{u}^{2}}{{\sin }^{2}}\theta /2g}{{{u}^{2}}{{\sin }^{2}}({{90}^{o}}-\theta )2g}$ $=\frac{{{\sin }^{2}}\theta }{{{\cos }^{2}}\theta }$ $={{\tan }^{2}}\theta$ $\therefore \,\,\,\,\,\frac{{{H}_{1}}}{{{H}_{2}}}={{\tan }^{2}}\theta$ Sample problem based on maximum height Problem 49. A cricketer can throw a ball to a maximum horizontal distance of 100 m. With the same effort, he throws the ball vertically upwards.  The maximum height attained by the ball is        [UPSEAT 2002] (a) 100 m          (b) 80 m            (c) 60 m            (d) 50 m Solution: (d) ${{R}_{\max }}=\frac{{{u}^{2}}}{g}=100\,m$                        (when $\theta =45{}^\circ$) $\therefore \,\,\,\,\,{{u}^{2}}=100\times 10=1000$ ${{H}_{\max }}=\frac{{{u}^{2}}}{2g}=\frac{1000}{2\times 10}=50\,\,metre.$            (when $\theta =90{}^\circ$) Problem 50. A ball thrown by one player reaches the other in 2 sec.  the maximum height attained by the ball above the point of projection will be about [Pb. PMT 2002] (a) 10 m            (b) 7.5 m           (c) 5 m              (d) 2.5 m Solution: (c) $T=\frac{2u\sin \theta }{g}=2\,sec$              (given) $\therefore \,\,\,\,\,u\sin \theta =10$ Now    $H=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}=\frac{{{(10)}^{2}}}{2\times 10}=5\,m.$ Problem 51. Two stones are projected with the same magnitude of velocity, but making different angles with horizontal. The angle of projection of one is p/3 and its maximum height is Y, the maximum height attained by the other stone with as p/6 angle of projection is [J & K CET 2000] (a) Y                 (b) 2 Y             (c) 3 Y               (d) $\frac{Y}{3}$ Solution: (d) When two stones are projected with same velocity then for complementary angles $\theta$ and $(9{{0}^{o}}\theta )$ Ratio of maximum heights: $\frac{{{H}_{1}}}{{{H}_{2}}}={{\tan }^{2}}\theta ={{\tan }^{2}}\frac{\pi }{3}=3\Rightarrow {{H}_{2}}=\frac{{{H}_{1}}}{3}=\frac{Y}{3}$ Problem 52. If the initial velocity of a projectile be doubled. Keeping the angle of projection same, the maximum height reached by it will (a) Remain the same       (b) Be doubled                (c) Be quadrupled           (d) Be halved Solution: (c) $H=\frac{{{u}^{2}}\sin 2\theta }{2g}$   $\therefore \,\,\,\,H\propto {{u}^{2}}$       [As $\theta =cons\tan t$] If initial velocity of a projectile be doubled then H will becomes 4 times. Problem 53. Pankaj and Sudhir are playing with two different balls of masses m and 2m respectively. If Pankaj throws his ball vertically up and Sudhir at an angle $\theta$, both of them stay in our view for the same period. The height attained by the two balls are in the ratio (a) 2 : 1             (b) 1 : 1             (c) $1:cos\,\theta$         (d) $1:sec\,\theta$ Solution: (b) Time of flight for the ball thrown by Pankaj ${{T}_{1}}=\frac{2{{u}_{1}}}{g}$ Time of flight for the ball thrown by Sudhir ${{T}_{2}}=\frac{2{{u}_{2}}\sin ({{90}^{o}}-\theta )}{g}=\frac{2{{u}_{2}}\cos \theta }{g}$ According to problem ${{T}_{1}}={{T}_{2}}\Rightarrow \frac{2{{u}_{1}}}{g}=\frac{2{{u}_{2}}\cos \theta }{g}\Rightarrow {{u}_{1}}={{u}_{2}}\cos \theta$ Height of the ball thrown by Pankaj ${{H}_{1}}=\frac{u_{1}^{2}}{2g}$ Height of the ball thrown by Sudhir ${{H}_{2}}=\frac{u_{2}^{2}{{\sin }^{2}}({{90}^{o}}-\theta )}{2g}$$=\frac{u_{2}^{2}{{\cos }^{2}}\theta }{2g}$ \ $\frac{{{H}_{1}}}{{{H}_{2}}}=\frac{u_{1}^{2}/2g}{u_{2}^{2}{{\cos }^{2}}\theta /2g}$ = 1           [As ${{u}_{1}}={{u}_{2}}\cos \theta$] Short Trick: Maximum height H µ T2 $\frac{{{H}_{1}}}{{{H}_{2}}}={{\left( \frac{{{T}_{1}}}{{{T}_{2}}} \right)}^{2}}$. \ $\frac{{{H}_{1}}}{{{H}_{2}}}=1$   (As ${{T}_{1}}={{T}_{2}}$) Problem 54. A boy aims a gun at a bird from a point, at a horizontal distance of 100 m. If the gun can impart a velocity of $500 m{{s}^{1}}$ to the bullet. At what height above the bird must he aim his gun in order to hit it (take $\operatorname{g} = 10 m{{s}^{2}}$) [CPMT 1996] (a) 20 cm           (b) 10 cm           (c) 50 cm           (d) 100 cm Solution: (a) Time taken by bullet to travel a horizontal distance of 100 m is given by $t=\frac{100}{500}=\frac{1}{5}sec$ In this time the bullet also moves downward due to gravity its vertical displacement $h=\frac{1}{2}g\,{{t}^{2}}=\frac{1}{2}\times 10\times {{\left( \frac{1}{5} \right)}^{2}}$$=1/5\,m= 20 cm$ So bullet should be fired aiming 20 cm above the bird to hit it. Problem 55. The maximum horizontal range of a projectile is 400 m. The maximum height attained by it will be  (a) 100 m          (b) 200 m          (c) 400 m           (d) 800 m Solution: (a) ${{R}_{\max }}=400\,m$ [when $\theta =45{}^\circ$] So from the Relation $R=4H\cot \theta \Rightarrow 400=4H\cot 45{}^\circ \Rightarrow \,H=100\,m.$ Problem 56. Two bodies are projected with the same velocity. If one is projected at an angle of ${{30}^{o}}$ and the other at an angle of ${{60}^{o}}$ to the horizontal, the ratio of the maximum heights reached is [EAMCET (Med.) 1995; Pb. PMT 2000; AIIMS 2001] (a) 3 : 1             (b) 1 : 3             (c) 1 : 2             (d) 2 : 1 Solution: (b) $\frac{{{H}_{1}}}{{{H}_{2}}}=\frac{{{\sin }^{2}}{{\theta }_{1}}}{{{\sin }^{2}}{{\theta }_{2}}}=\frac{{{\sin }^{2}}{{30}^{o}}}{{{\sin }^{2}}{{60}^{o}}}=\frac{1}{3}$ Problem 57. If time of flight of a projectile is 10 seconds. Range is 500 m. The maximum height attained by it will be [RPMT 1997; RPET 1998] (a) 125 m          (b) 50 m            (c) 100 m           (d) 150 m Solution: (a) $T=\frac{2u\sin \theta }{g}=10\,\sec \Rightarrow \,u\sin \theta =50$ so $H=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}=\frac{{{(50)}^{2}}}{2\times 10}=125\,m$. Problem 58. A man can throw a stone 80 m. The maximum height to which he can raise the stone is (a) 10 m            (b) 15 m            (c) 30 m             (d) 40 m Solution: (d) The problem is different from problem no. (54). In that problem for a given angle of projection range was given and we had find maximum height for that angle. But in this problem angle of projection can vary, ${{R}_{\max }}=\frac{{{u}^{2}}}{g}=80\,m$  [for $\theta =45{}^\circ$] But height can be maximum when body projected vertically up ${{H}_{\max }}=\frac{{{u}^{2}}{{\sin }^{2}}{{90}^{o}}}{2g}=\frac{{{u}^{2}}}{2g}=\frac{1}{2}\left( \frac{{{u}^{2}}}{g} \right)=40\,m$ Problem 59. A ball is thrown at different angles with the same speed u and from the same points and it has same range in both the cases. If ${{y}_{1}}and\text{ }{{y}_{2}}$ be the heights attained in the two cases, then ${{y}_{1}}+{{y}_{2}}=$ (a) $\frac{{{u}^{2}}}{g}$                   (b) $\frac{2{{u}^{2}}}{g}$      (c) $\frac{{{u}^{2}}}{2g}$                  (d) $\frac{{{u}^{2}}}{4g}$ Solution (c) Same ranges can be obtained for complementary angles i.e. $\theta$ and ${{90}^{o}}\theta$ ${{y}_{1}}=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}$ and ${{y}_{2}}=\frac{{{u}^{2}}{{\cos }^{2}}\theta }{2g}$     $\therefore \,\,\,\,{{y}_{1}}+{{y}_{2}}=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}+\frac{{{u}^{2}}{{\cos }^{2}}\theta }{2g}=\frac{{{u}^{2}}}{2g}$

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