Question

A ball is thrown upwards with a velocity of 29ms

-1

in the absence of

air resistance. A) Mention the direction of the acceleration during

the upward motion of the ball. B) What is the velocity and

acceleration of the ball at the highest point? C) How much high the

ball goes up? D) Find the time taken by the ball to reach the ground. g=9.8ms

-2

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Answer 1

Explanation:

A) The Direction of acceleration is towards the earth's centre.

B) Velocity at the highest point is 0 and acceleration is also 0

C) By the 3rd equation of motion

v = final velocity

u = initial velocity

a = g = 9.8m/s^2

s = distance

$v ^{2} = u {}^{2} + 2as$

u = 29m/s

v = 0

a = -9.8m/s^2

s = ?

Therefore,

0 = 29^2 + 2(-9.8)s

0 = 841 - 19.6s

Therefore,

-841 = -19.6s

841 = 19.6s

s = 841 / 19.6

s = 42.9m

D) By second equation of motion

$v = u + at$

v = 29m/s.............(because after getting top u becomes 0 and v becomes the velocity at which the object is thrown. This works only if air resistance is not taken in account.)

u = 0

a = g = 9.8m/s^2

t = ?

Therefore,

29 = 0 + 9.8t

29 = 9.8t

t = 29 / 9.8

t ≈ 2.95seconds

Hope this helps