For an observer on trolley, the direction of projection of particle is shown in figure, while for observer on ground ball rises vertically. Maximum height (in meter) seen from the trolley is:
(take $g = 10 {ms}^{- 2}$ )

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Solution

Here $v_{H} = 0$
$v_{H} =$ Horizontal velocity
So, $- v c o s 60^{\circ} + 10 = 0 v \times \frac{1}{2} = 10 v = 20 m / s e c .$

$v_{y}$ = Vertical velocity
$v_{y} = v s i n 60^{\circ} v_{y} = 20 \times \frac{\sqrt{3}}{2} = 10 \sqrt{3} m / s e c$
Using 3rd equation of motion-
$v^{2} = u^{2} + 2 a s 2 a h = v^{2} - u^{2}$
$2 \times (- 10) \times h = 0 - (10 \sqrt{3})^{2} - 20 h = - 300 h = \frac{300}{20} = 15 m$

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For an observer on trolley, the direction of projection of particle is shown in figure, while for observer on ground ball rises vertically. Maximum height (in meter) seen from the trolley is: (take g=10 ms−2)

Here vH=0 vH= Horizontal velocity So, −v cos 60∘+10=0v×12=10v=20 m/sec. vy = Vertical velocity vy=v sin 60∘vy=20×√32=10√3 m/sec Using 3rd equation of motion- v2=u2+2as2ah=v2−u2 2×(−10)×h=0−(10√3)2−20h=−300h=30020=15 m

For an observer on trolley, the direction of projection of particle is shown in figure, while for observer on ground ball rises vertically. Maximum height (in meter) seen from the trolley is:
(take $g = 10 {ms}^{- 2}$ )