Question

If a ball is thrown at angle 30° from a tower of height 20m with initial velocity 30m/s.
(i) Find the time taken to reach the ground
(ii) Find horizontal range.
(iii) Find velocity just before hitting ground.

Answer 1

Answer:

A ball is thrown from the top of a tower with an initial velocity of 10 ms  

−1

 at an angle of 30  

∘

 with the horizontal If it hits the ground at a distance of 17.3 m from the base of the tower, the height of the tower is (Take g= 10 ms−2)

Explanation:

The ball is thrown at an angle, θ=30  

o

.

Initial velocity of the ball, u=10 m/s  

Horizontal range of the ball, R=17.3 m

We know that, R=u cosθ t,

where t is the time of flight  

⟹t=  

5  

3

​

 

17.3

​

=2 secs

using equation of motion we get:-

−h=ut−  

2

1

​

gt  

2

 

=10×2−  

2

1

​

10×2  

2

=−10

⟹ Height of tower, h=10 m