Question

on the moon and on
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.
14
A stone is released from the toneftower of height 196 m​

Answer 1

Answer:

Maximum height = 122.5 m, Total time taken = 10 s

Explanation:

$\red{\boxed{\rm Solution}}$

Given,

u = 49 m/s

g = 9.8 m/s²

a = - g = - 9.8 m/s²

i) To find,

The maximum height to which it rises

When a body is vertically thrown upwards then at its maximum height it's velocity becomes 0 m/s and after that due to gravitional force it starts to fall down.

Therefore from the above statement,

v = 0 m/s at maximum height

Applying the third equation of motion, i.e v² - u² = 2as

0 - (49)² = 2h(-9.8) [h is the maximum height to which the body rises]

0 - 2401 = -19.6h

-2401 = -19.6h

2401 = 19.6h

h = 2401/19.6 = 122.5 m

$\boxed{\rm Maximum\:height\:achieved\:by\:the\:body\:=\:122.5m}$

ii) When a body is vertically thrown upwards then the time taken to reach the maximum height (time of ascent) is equal to the the time taken to reach the surface of the earth from the maximum height (time of descent)

From the above statement,

Time of ascent = Time of descent

We know the maximum height of the body, i.e 122.5 m

We can calculate the time taken to reach the maximum height, i.e time of ascent by using first equation of motion

At maximum height v = 0 m/s

v = u + at

0 = 49 - 9.8t

49 = 9.8t

t = 5 s

Total time taken to return to the surface = Time of ascent + Time of descent = 5 + 5 = 10 s

$\boxed{\rm Time\:taken\:by\:the\:body\:to\:return\:to\:surface=\:10s}$

Equations of motion:-

1. v = u + at
2. s = ut + (at²/2)
3. v² - u² = 2as

Important things to remember while solving questions related to free fall:-

1. At the maximum height the velocity of the body becomes 0 m/s
2. Time of ascent = Time of descent
3. Maximum height = u²/2g
4. Time of ascent = Time of descent = u/g
5. Time of flight = 2u/g