Question

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

Answer 1

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,

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

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

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

Answer 2

Given :  A ball kicked from ground level at an initial velocity of 60 m/s and an angel with ground reaches a horizontal distance of 200 meters.

To Find :  size of  angle

Solution:

Let say angle α

Horizontal Speed = 60 cosα

Vertical Speed = 60 sinα

V = U + at

a = - g = - 19

t = time to reach max height

=> 0 = 60 sinα - (10)t

=> t = 6sinα

time of flight = 2t  = 12sinα

horizontal distance = 60 cosα  * time of flight

=  60 cosα  *   12sinα

= 720  sinαcosα

= 720 sin2α   / 2

= 360 sin2α

360 sin2α = 200

=> sin2α = 200/360

=> sin2α = 5/9

=> 2α = sin⁻¹(5/9)

=>  2α =   33.749°

=> 2α =   33.749°

      α =   16.875°

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