Question

A boy throws a ball with a speed of 50 m/s in upwards direction. It falls back back after 5 sec Find hight when a = 5m/sec square ​

Answer 1

Answer:

Using equations of motion:

S=ut+1/2 at²

S=50*5+1/2*5*5²

S=250+125/2

S=250+62.5

S=312.5m

So the height when a=5m/sec is 312.5 m

Explanation:

Answer 2

Given :-

• Initial velocity, u = 50 m/s

• Final velocity, v = 0 m/s

( At the maximum height)

• Acceleration, a = -5 m/s²

( Negative sign because is going upward)

To Find :-

• Height

Solution :-

To find height, we have to use 3rd equation of motion.

We know,

$\sf v^2-u^2=2as$

Where,

v = Final velocity

u = Initial velocity

s = Height / Distance covered

a = Acceleration

Now, put the given values

⟹ 2as = v² - u²

⟹ 2as = (0)² - ( 50)²

⟹ 2 × -5 × s = - 2500

⟹ s = 2500 / 10

⟹ s = 250 m

Hence,

the maximum height attained by the ball is

250 m.