Question

A person throws a ball upwards with some speed so that it travels a distance of 25 meter before it

stops and starts descending towards the earth. The ball returns to the same spot on the ground. It is

known from an approximation that the acceleration due to gravity in the area is 10 m/s2

in the area.

The ball takes the same time to go up and to come down. Based on the information provided and

using the equations of motion, your friend was asked to solve a number of questions. However your

friend found the following questions tough to answer, please help your friend to answer the following

questions:

a) How long did the ball take to travel to the maximum height?

b) What was the speed with which the ball was thrown up?

c) Why did the ball take the same time to travel up and come down?

d) If this experiment was done while standing on a truck moving at a uniform speed, will it give us

the same result? Explain briefly.​

Answer 1

Given:-

$g \: = 10 \: m { s}^{ - 2}$

$h = 25 \: m$

(a)

to find time taken to reach max height...

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

final velocity become zero,as ball cones to rest so.,

${0}^{2} = {u}^{2} + 2 \times ( - 10 )\times 25$

${u}^{2} = 500$

$u = \sqrt{500}$

now,

$v = u + gt$

final velocity become zero as ball comes to rest

so.,

$0 = u + ( - 10) \times t$

$t = \frac{ - u}{ - 10} or \: \frac{u}{10}$

putting value of u.,

$t \: = \frac{ \sqrt{500} }{10} s \: or \: \sqrt{5} s$

(b)

$the \: initial \: speed \: = u = \sqrt{500} m {s}^{ - 1} \: or \: 10 \sqrt{5} m {s}^{ - 1}$

(c)

The time required for initially horizontal projectile motion to occur is the same as the time required for the object to fall to its final height. Thus, a ball thrown horizontally will reach the ground at the same time as a ball dropped from the same height.

(d)

yes,

The reason is that at the moment the ball was thrown, the ball was in motion along with the person and the truck, due to the inertia of motion. So during the time ball remains in air, both the person and the ball move ahead by the same distance.