Question

 A ball is thrown vertically upwards rises to a height of 50m. If the acceleration due to gravity g = 9.8 m/s², calculate a) the initial velocity with which the ball was thrown. b) the time taken to attain the maximum height​

Answer 1

Given :

A ball is thrown vertically upward rises to a height of 50m.

To Find :

• Initial velocity with which the ball was thrown.
• Time taken by ball to attain the maximum height.

Solution :

❖ For a body thrown vertically upward, g is taken negative.

Since constant acceleration due to gravity acts on the ball throughout the motion, equation of kinematics can be easily applied to solve this question$.$

A] Initial velocity of ball :

Applying 3rd equation of kinematics;

➙ v² - u² = 2gH

• At max. height, v = zero

➙ 0² - u² = 2(-9.8)(50)

➙ -u² = -980

➙ u = 31.3 m/s

B] Time of ascent :

Applying 1st equation of kinematics;

➙ v = u + gt

➙ 0 = 31.3 + (-9.8)t

➙ -31.3 = -9.8t

➙ t = 31.3/9.8

➙ t = 3.19 s

Answer 2

Qᴜᴇsᴛɪᴏɴ:-

➨ ᴀ ʙᴀʟʟ ɪs ᴛʜʀᴏᴡɴ ᴠᴇʀᴛɪᴄᴀʟʟʏ ᴜᴘᴡᴀʀᴅ ʀɪsᴇs ᴛᴏ ᴀ ʜᴇɪɢʜᴛ ᴏғ 50ᴍ. ɪғ ᴛʜᴇ ᴀᴄᴄᴇʟᴀʀᴀᴛɪᴏɴ ᴅᴜᴇ ᴛᴏ ɢʀᴀᴠɪᴛʏ ɢ=9.8ᴍ/s²,ᴄᴀʟᴄᴜʟᴀᴛᴇ ᴀ)ᴛʜᴇ ɪɴɪᴛɪᴀʟ

ᴠᴇʟᴏᴄɪᴛʏ ᴡɪᴛʜ ᴇʜɪᴄʜ ᴛʜᴇ ʙᴀʟʟ ᴡᴀs ᴛʜʀᴏᴡɴ.

ʙ)ᴛʜᴇ ᴛɪᴍᴇ ᴛᴀᴋᴇɴ ᴛᴏ ᴀᴛᴛᴀɪɴ ᴛʜᴇ ᴍᴀxɪᴍᴜᴍ ʜᴇɪɢʜᴛ.

Aɴsᴡᴇʀ:-

➨ᴛᴏ ғɪɴᴅ:-

ɪɴɪᴛɪᴀʟ ᴠᴇʟᴏᴄɪᴛʏ ᴀɴᴅ ᴛʜᴇ ᴛɪᴍᴇ ᴛᴀᴋᴇɴ

➨ɪɴɪᴛɪᴀʟ ᴠᴇʟᴏᴄɪᴛʏ

ʙʏ ᴀᴘᴘʟʏɪɴɢ 3ʀᴅᴇǫᴜᴀᴛɪᴏɴ ᴏғ ᴍᴏᴛɪᴏɴ.

ᴠ²-ᴜ²=2ɢʜ

ɪɴ ᴡʜɪᴄʜ ᴠ=0

0²-ᴜ²=2×(-9.8)×50

ᴜ²=980

ᴜ=31.304ᴍ/s

➨ᴛɪᴍᴇ ᴛᴀᴋᴇɴ

ʙʏ ᴀᴘᴘʟʏɪɴɢ 1sᴛ ᴇǫᴜᴀᴛɪᴏɴ ᴏғ ᴍᴏᴛɪᴏɴ.

ᴠ=ᴜ+ɢᴛ

0=31.304+(-9.8)ᴛ

ᴛ= 31.304/9.8

ᴛ=3.1942s

sᴏ ᴛʜᴇ ɪɴɪᴛɪᴀʟ ᴠᴇʟᴏᴄɪᴛʏ ɪs 31.304ᴍ/s ᴀɴᴅ ᴛɪᴍᴇ ɪs 3.1942s

ғᴏʀᴍᴜʟᴀ ᴜsᴇᴅ:-

➨ᴠ²-ᴜ²=2ɢʜ

➨ᴠ=ᴜ+ɢᴛ

ᴀɴᴅ ᴀʟʟ ᴡᴇ ᴀʀᴇ ᴅᴏɴᴇ✔️

ʜᴏᴘᴇ ɪᴛs ʜᴇʟᴘs ᴜʜ❣️