Question

A ball is thrown vertically upwards with an initial velocity of 19.6 m/s. It returns 4s later. The greatest height reached by ball will be ? Take g = ms-2

Answer 1

Answer:

Explanation:

(i) 45 m, (ii) 30 m s  

−1

 

The ball is thrown up and it returns in 6sec,  

Time of accent = Time of decent

So, time to reach the highest point =6/2=3s

By 2nd equation of motion

s=ut+  

2

1

​

gt  

2

 

To calculate height, consider motion of the ball from highest point to the ground

H=  

2

1

​

gt  

2

=  

2

1

​

×10×3  

2

=45 m

To calculate projection velocity, consider motion of the ball from projection to return to the point of projection.

0=ut−  

2

1

​

gt  

2

 

u=10×6/2=30 m/s