Question

13. A ball is thrown vertically upwards with a velocity of 49 m/s.
Calculate
(i) the maximum height to which it rises,
(ii) the total time it takes to return to the surface of the earth.​

Answer 1

Explanation:

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Answer 2

Answer:

Explanation:

According to the equation of motion under gravity:

v2 - u2 = 2gs

Where,

u = Initial velocity of the ball

v = Final velocity of the ball

s = Height achieved by the ball

g = Acceleration due to gravity

At maximum height, final velocity of the ball is zero, i.e., v= 0

u = 49 m/s

During upward motion, g = -9.8 m s-2

Let h be the maximum height attained by the ball.

Hence,

0 - 492 = 2×9.8×h

⇒ h = 492/(2×9.8)

⇒ h = 2401/19.6 = 122.5

Let t be the time taken by the ball to reach the height 122.5 m, then according to the equation of motion:

v = u + gt

We get,

0 = 49 + t x (- 9.8)

9.8t = 49

t = 49 / 9.8 = 5s

Also,

Time of ascent = Time of descent

Therefore, total time taken by the ball to return = 5 + 5 = 10 s