Question

A ball is thrown vertically upward with initial velocity of 30 ms-1 . taking g= 10 ms-2, find the maximum height reached by the stone. what is the total distance covered by the ball before it reaches the ground?

Answer 1

Answer:-

Given:

Initial Velocity of the ball (u) = 30 m/s

acceleration due to gravity (g) = 10 m/s²

Here , the ball is thrown in the opposite direction of the earth , so value of g will be negative.

⟶ g = - 10 m/s²

Final velocity (v) = 0 m/s [ it comes to rest when it reaches the maximum height ]

we can find the maximum height reached by the ball using the third equation of motion.

i.e.,

v² - u² = 2aS

here , a = g & S = height reached (say h )

So,

⟶ v² - u² = 2gh

Putting the values we get,

⟶ 0² - (30)² = 2 ( - 10)(h)

⟶ - 900 = - 20h

⟶ - 900 / - 20 = h

⟶ 45 m = h

Now,

Total distance covered = 2h (height reached by the ball = distance covered to reach the ground)

So,

Total distance covered = 2(45) = 90 m

∴

• Maximum height reached by the ball is 45 m
• Total distance covered by the ball = 90 m.