First slide

Projection Under uniform Acceleration

Question

A cricketer of height 2.5 m throws a ball at an angle 30º with the horizontal such that it is received by another cricketer of same height standing at a distance of 50 m from the first one.The maximum height attained by the ball from the ground is

Moderate

A

3.6 m
B

6.1 m
C

7.2 m
D

9.7 m

Solution

R = 50 m
$\begin{array}{l} \frac{H}{R} = \frac{1}{4} \tan θ \\ H = \frac{50}{4} \frac{1}{\sqrt{3}} = 7 .2 m \end{array}$
from the ground height is 7.2 + 2.5 = 9.7 m

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