JEE Main · 2026 · 24 Jan 2026 · Shift 1 · Kinematics

A boy thrown a ball into air at

4 5^{\circ}

from the horizontal to land it on a roof of a building of height H. If the ball attains maximum height in 2 s and lands on the building in 3 s after launch, then value of H is ______ m.
(

g = 10 m / s^{2}

)

(A)20
(B)10
(C)25
(D)15✓ Correct

Answer: (D)

Solution

The time to reach maximum height is the time for the vertical velocity to vanish, so

u_{y} = g t_{m a x} = 10 \times 2 = 20 m/s

. The height of the point reached

t = 3

s after launch is

H = u_{y} t - \frac{1}{2} g t^{2} = 20 \times 3 - \frac{1}{2} \times 10 \times 3^{2} = 60 - 45

. The building height is

15

JEE Main · 2026 · 24 Jan 2026 · Shift 1 · Single correct option

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