Question

A ball is dropped from height 2.50m above the floor .find the speed v with which it reaches the floor . The ball now rebounds . The speed of the ball is decreased to 3v/4 due to this collision . How high will the ball rise ?

Answer 1

1.
v = sqrt(2gH)
= sqrt(2 × 9.8 × 2.5)
= 7 m/s
It hits the ground with speed equal to 7 m/s

2.
Kinetic energy of ball after losing some energy due to collision is
K = 0.5mv^2
= 0.5m × (3v/4)^2
= 0.5m × (3 × 7/4)^2
= 0.5m × (21 / 4)^2

This energy is converted into potential energy at height “h”
0.5m × (21 / 4)^2 = mgh
h = (21 / 4)^2 / (2g)
h = (21 / 4)^2 / (2 × 9.8)
h = 1.40 m

The ball will rise to the height equal to 1.40 m

Answer 2

╭━─━─━─━─≪✠≫─━─━─━─━╮
✮❀ӇЄƦЄ ƖƧ ƳƠƲƦ ƛƝƧƜЄƦ❀✮
╰━─━─━─━─≪✠≫─━─━─━─━╯

[➊]✮✮✮✮✮

Vo ➙ 0 m/s
H ➙ 2.5 m
g ➙ 9.8 m/s^2

V ➙ √(2gH)
...➙ √(2 × 9.8 × 2.5)
...➙ 7 m/s

[➋] ✮✮✮✮✮

(Kɪɴᴇᴛɪᴄ ᴇɴᴇʀɢʏ ᴏғ ʙᴀʟʟ ᴀғᴛᴇʀ ʟᴏꜱɪɴɢ ᴛʜᴇ ꜱᴏᴍᴇ ᴇɴᴇʀɢʏ ᴅᴜᴇ ᴛᴏ ᴄᴏʟʟɪꜱɪᴏɴ ᴀʀᴇ)

K➙ 0.5 mv^2

..➙ 0.5m × (3√/4)^2

..➙ 0.5m × (3 × 7/4)^2

..➙ 0.5m × (21 /4 )^2

(Tʜᴇɴ, Tʜɪꜱ ᴇɴᴇʀɢʏ ɪꜱ ᴄᴏɴᴠᴇʀᴛᴇᴅ ɪɴᴛᴏ ᴘᴏᴛᴇɴᴛɪᴀʟ ᴇɴᴇʀɢʏ ᴀᴛ ʜᴇɪɢʜᴛ 'ʜ')

....➙ 0.5m × (21 /4 )^2 = mgh

H ➙ (21 /4 )^2 / (2g)

H ➙ (21 /4 )^2 / (2 × 9.8)

H ➙ 1.40 m...✔

Sᴏ, ᴛʜᴇ ʙᴀʟʟ ᴡɪʟʟ ʀɪꜱᴇ ᴛᴏ ᴛʜᴇ ʜᴇɪɢʜᴛ ɪꜱ ᴇϙᴜᴀʟ ᴛᴏ 1.40m.