Question

A ball kicked from ground level at an initial velocity of 60 m/s and an angle θ with ground reaches a horizontal distance of 200 meters. 
a) What is the size of angle θ? 
b) What is time of flight of the ball?​

Answer 1

Answer:

Please check the attached pic for solution

Answer 2

Answer:

a) The size of the angle θ is 16.9°.

b) The time of flight of the ball is 3.48 seconds.

Explanation:

Given:

• Initial velocity, u = 60 m/s.
• Range, R = 200m.

To Find:

a) What is the size of angle θ ?

b) What is time of flight of the ball ?

Solution:

To find the angle θ = ?

(a) We know that,

$\qquad \quad\sf R = \dfrac{u^2 sin 2 \theta}{g}$

$\qquad : \implies 200m = \dfrac{60^2 (sin2 \theta)}{10}$

$\qquad \quad : \implies sin 2 \theta = \dfrac{2000}{3000}$

$\qquad \qquad : \implies \sf sin 2 \theta = \dfrac{5}{9}$

$\qquad \quad : \implies \sf 2 \theta = sin^{-1} \bigg( \dfrac{5}{9} \bigg)$

$\qquad \quad : \implies \sf \dfrac{1}{2} (33.8 ^{\circ}$

$\qquad \qquad \large \dag{\underline{\boxed{\sf \theta = 16.9 ^{\circ}}}}$

Now,

To find the time of flight = ?

(b) Time of flight of the ball.

We know that,

$\qquad \qquad \sf T = \dfrac{2u sin \theta}{g}$

$\qquad \sf \dfrac{2 \times 60 \times sin (16.9^{\circ})}{10}$

$\qquad \qquad \large \dag{\underline{\boxed{\sf T = 3.48 sec}}}$