First slide

Projection Under uniform Acceleration

Question

A cricketer hits a ball with a velocity 25 m/s at 60^o above the horizontal. How far above the ground it passes over a fielder 50m from the bat (assume the ball is struck very close to the ground)

Moderate

A

12.7 m
B

11.6 m
C

9.0 m
D

8.2 m

Solution

The horizontal component of velocity is given by
$V_{X} = 25 \cos 60^{\circ} = 12 \cdot 5 m / s$
Similarly, the vertical component of velocity is given by
$v_{y} = 25 \sin 60^{\circ} = 12 \cdot 5 \times \sqrt{3} m / s$
Time to cover 50 metre distance
$t = \frac{50}{12 \cdot 5} = 4 \sec$
The vertical height x is given by (see fig. 10)
$\begin{array}{l} x = V_{x} t - \frac{1}{2} \bar{g} t^{2} \\ = 12 \cdot 5 \sqrt{\bar{3}} \times 4 - \frac{1}{2} \times 9 \cdot 8 \times 16 \\ = 8 \cdot 2 m \end{array}$

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