A Ball is thrown vertically upwards – physics numerical

In this post, we have solved numerical problems related to Vertical Motion, specifically when a ball is thrown vertically upwards.

A Ball is thrown vertically upwards – physics numerical (solved)

Q1) A ball is thrown vertically upwards with a velocity of 49 m/s. Calculate (1) The maximum height to which it rises. (2) The total time it takes to return to the surface of the earth.

Solution:

Problem-Data given

• Initial velocity of the ball = u = 49m/s.
• The velocity of the ball at maximum height = v = 0.
• g = 9.8m/s²

We have to Find out

1) The maximum height to which it rises.

2) The total time it takes to return to the surface of the earth.

Solving the problem:

Say, the time to reach the maximum height H is t

Consider a formula for upward motion with retardation,

v² = u² – 2gH

=> 0 = (49)² – 2 × 9.8 × H

2 × 9.8 × H = (49)²

19.6 H = 2401

H = 122.5 m

Now consider the below-mentioned formula for the upward motion with retardation,

v = u – g × t

=> 0 = 49 – 9.8 × t

=> 9.8t = 49

t = 5 sec

Answer:

1) The maximum height to which it rises: H = 122.5 m

2) The total time it takes to return to the surface of the earth: t= 5 sec

A Ball is thrown vertically upward – physics numerical (solved)

Q2) When a ball is thrown vertically upwards, it goes through a distance of 19.6 m. Find the initial velocity of the ball and the time taken by it to rise to the highest point. (Acceleration due to gravity, g=9.8 ms^-2)

Solution:

Distance traveled =s=19.6 m

We know that the velocity at the highest v = 0

say, initial velocity = u m/s & the time taken to rise to the highest point = t sec

Using the following equation of motion with retardation, we can find the initial velocity = u

v²=u²– 2gs

=> 0 = u² – 2 x 9.8 x 19.6

=> u²=19.6 x 19.6

u = 19.6 m upwards

Now we will find out the the time taken to rise to the highest point (t):

v = u – gt

=> 0 = u – gt

=> t = u/g = 19.6/9.8 s = 2 seconds

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