Question

A cricket ball is thrown at a speed of 28 m s(-1) in a direction of 30° above the horizontal. Calculate (a) the maximum height, (b) the time taken by the ball to return to the same level, and (c) the distance from the thrower to the point where the ball returns to the same level.​

Answer 1

$\underline{\huge{Answer:-}}$

Explanation:

(a) The maximum height is given by

$\large \mathcal{ h_{m} = \frac{ (v_{o} sin \emptyset _{o} ) {}^{2} }{2g} = \frac{(28sin \: 30 \degree) {}^{2} }{2(9.8)}m }$

$\large \mathcal {\frac{14 \times 14}{2 \times 9.8} = 10.0m}$

(b) The time is taken to return to the same level is

$\large \mathcal{ T_{f} = (2 \: v_{o} \: sin \: \emptyset _{o}) /g = (2 \times 28 \times sin \: 30 \degree)/9.8 }$

$\large \mathcal{ = 28/9.8 \: s = 2.9s}$

(c) The distance from the thrower to the point where the ball returns to the same level is

$\large \mathcal{R = \frac{ v_{o} {}^{2} sin \: 2 \emptyset _{o} }{g} = \frac{28 \times 28 \times sin \: 60 \degree}{9.8} = 69m}$

Answer 2

hilu ❤️

Solution :

Given, speed u=28ms−1u=28ms-1 and θ=30∘θ=30∘

∴∴ The time taken by the ball to return the same level is

T=2u/sinθg

=2×28×sin30∘/ 9.8

=28/9.8

=2.9s

mark ❤️