Question

A person throws a ball in the upward direction with a speed of 5 m/s. And the ball reaches back to the person in 10 seconds. Find the maximum height attained by the ball.

Answer 1

Answer:

Explanation:

time of flight (t) = 2 x initial velocity (v) / acceleration due to gravity (g)

t = 2v/g

10 = 2v/g

10 = 20/g

g = 2 m/$s^{2}$

By conservation of energy:

E_i = E_f

1/2 m $v^{2}$ =  mgh

h = $v^{2}$ /2g

h = 100/4

h = 25 m

Answer 2

Answer:

Explanation:

Initial velocity, u =  ?

Final velocity, v = 0

Acceleration due to gravity, g = 10 m/s  

2

 

Height, h = 5 m

Using relation, for a freely falling body:

v  

2

=u  

2

+2gh

(0)  

2

=(u)  

2

+2×(−10)×5

0  = u  

2

- 100

u  

2

=100

So, u = 10 m/s

(b) Using relation v = u + gt

0 = 10 + (-10) t

-10 = -10t

t = 1 sec