JEE Main · 2021 · 27 Aug 2021 · Shift 2 · Kinematics

A player kicks a football with an initial speed of

25 ms^{- 1}

at an angle of

4 5^{\circ}

from the ground. What are the maximum height and the time taken by the football to reach at the highest point during motion? (Take

g = 10 ms^{- 2}

)

(A) $h_{m a x} = 15.625 m, T = 3.54 s$
(B) $h_{m a x} = 10 m, T = 2.5 s$
(C) $h_{m a x} = 15.625 m, T = 1.77 s$ ✓ Correct
(D) $h_{m a x} = 3.54 m, T = 0.125 s$

Answer: (C)

Solution

The vertical component of the launch velocity is

u_{y} = u sin 4 5^{\circ} = \frac{25}{2}

m/s. The maximum height is

h_{m a x} = \frac{u _{y}^{2}}{2 g} = \frac{( 25/ 2 ) ^{2}}{2 \times 10} = \frac{312.5}{20} = 15.625 m,

and the time to reach it is

T = \frac{u _{y}}{g} = \frac{25/ 2}{10} = 1.77 s .

The football rises to

15.625

m and takes

1.77

s to reach the highest point.

JEE Main · 2021 · 27 Aug 2021 · Shift 2 · Single correct option

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