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 10 m from the ground? (g=10 m/s2, sin 30o=1/2, cos 30o=√3/2)

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 10 m from the ground? (g=10 m/s2, sin 30o=1/2, cos 30o=√3/2)  (A) 5.20m (B) 4.33m (C) 2.60m (D) 8.66m

Tardigrade Admin · Accepted Answer

The ball will be at point P when it is at a height of 10 m from the ground. So, we have to find distance OP , which can be calculated direct considering it as a projectile on a levelled (OX) . OP=R=(u2 sin 2θ /g) =(102× sin (2× 30o)/10) =(10√3/2)=5√3 =8.66 m

Q. 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 10 m from the ground? (g=10m/s2,sin30o=1/2,cos30o=3​/2)

5.20m

4.33m

2.60m

8.66m

The ball will be at point P when it is at a height of 10m from the ground. So, we have to find distance OP , which can be calculated direct considering it as a projectile on a levelled (OX) . OP=R=gu2sin2θ​ =10102×sin(2×30o)​ =2103​​=53​ =8.66m

Q. 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 $30^{\circ}$ with the horizontal. How far from the throwing point will the ball be at the height of 10 m from the ground? $(g = 10 m / s^{2}, sin 30^{o} = 1/2, cos 30^{o} = 3 /2)$