Kinematics

JEE Main level, single correct, medium

202 of 308

Question 202 of 308

A boy throws a ball into air at 45∘45^{\circ} from the horizontal to land it on a roof of a building of height HH. If the ball attains maximum height in 2 s and lands on the building in 3 s after launch, then value of HH is ____\_\_\_\_ m.
(g=10 m/s2)\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)

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Answer: (A) 15

For projectile motion at 45°, vertical component of initial velocity:
v0y=v0sin⁡45°=v02v_{0y} = v_0 \sin 45° = \frac{v_0}{\sqrt{2}}
Maximum height occurs at t=2t = 2 s when vy=0v_y = 0:
0=v0y−gt0 = v_{0y} - gt
⇒v0y=10×2=20\Rightarrow v_{0y} = 10 \times 2 = 20 m/s
Thus v02=20\frac{v_0}{\sqrt{2}} = 20, so v0=202v_0 = 20\sqrt{2} m/s
Height at landing time t=3t = 3 s:
H=v0yt−12gt2=20×3−12×10×9H = v_{0y}t - \frac{1}{2}gt^2 = 20 \times 3 - \frac{1}{2} \times 10 \times 9
=60−45=15= 60 - 45 = 15 m

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