Question

A boy playing on the roof of a 10 m high
building throws a ball with a speed of 10
m/s at an angle of 30° with the horizontal.
How far from the throwing point will the
ball be at the height of 10m from the
ground?
Tg = 10 m/s2, sin 30º = 5, cos 30º = 13/2]
(A) 4.33 m
(B) 2.60 m
(C) 8.66 m
(D) 5.20 m​

Answer 1

Answer:

c) 8.66 m

Explanation:

Answer 2

The distance covered by the ball at the height of 10 m from the  ground is 8.66 meters.

Explanation:

Given that,

Speed of the ball, u = 10 m/s

Angle of projection, $\theta=30^{\circ}$

We need to find the distance covered by the ball at the height of 10 m from the  ground. It means that we need to find the horizontal range of the projectile as :

$R=\dfrac{u^2\ sin2\theta}{g}$

$R=\dfrac{(10)^2\ sin2(30)}{10}$

R = 8.66 meters

So, the distance covered by the ball at the height of 10 m from the  ground is 8.66 meters. Hence, this is the required solution.

Learn more,

Projectile motion

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