Question

A ball thrown up vertically returns to the thrower after 6 s. Find
(a) the velocity with which it was thrown up,
(b) the maximum height it reaches, and
(c) its position after 4 s.

Answer 1

Answer:

ello

gud mrng

hru

here izz ur answer

(≡^∇^≡)

Time to reach Maximum height,

t = 6/2 = 3 s.

v = 0 (at the maximum height)

a = – 9.8 m s–²

Since the ball is thrown upwards, the acceleration is negative.

(i) Velocity

Using,

v = u + at, we get

0 = u – 9.8 × 3

or, u = 29.4 ms–¹

Thus, the velocity with which it was thrown up = 29.4ms–¹

(ii) Maximum height it reaches

Using, 2aS = v² – u², we get

S = v²- u²/2a

S = 0 – 29.4 × 29.4/(- 2× 9.8)

S = 44.1 m

Thus, Maximum height it reaches = 44.1 m.

(iii) Position after 4 seconds

t = 4s.

In 3 s, the ball reaches the maximum height and in 1 s it falls from the top.

Distance covered in 1 s from maximum height,

S = ut + 1/2at ²

S = 0 + 1/2 × 9.8 × 1

S = 4.9 m

Answer 2

ᴇʟᴏ

ᴛɪᴍᴇ ᴛᴏ ʀᴇᴀᴄʜ ᴍᴀxɪᴍᴜᴍ ʜᴇɪɢʜᴛ,

ᴛ = / = s.

ᴠ = (ᴀᴛ ᴛʜᴇ ᴍᴀxɪᴍᴜᴍ ʜᴇɪɢʜᴛ)

ᴀ = – . ᴍ s–²

sɪɴᴄᴇ ᴛʜᴇ ʙᴀʟʟ ɪs ᴛʜʀᴏᴡɴ ᴜᴘᴡᴀʀᴅs, ᴛʜᴇ ᴀᴄᴄᴇʟᴇʀᴀᴛɪᴏɴ ɪs ɴᴇɢᴀᴛɪᴠᴇ.

(ɪ) ᴠᴇʟᴏᴄɪᴛʏ

ᴜsɪɴɢ, ᴠ = ᴜ + ᴀᴛ, ᴡᴇ ɢᴇᴛ

= ᴜ – . ×

ᴏʀ, ᴜ = . ᴍs–¹

ᴛʜᴜs, ᴛʜᴇ ᴠᴇʟᴏᴄɪᴛʏ ᴡɪᴛʜ ᴡʜɪᴄʜ ɪᴛ ᴡᴀs ᴛʜʀᴏᴡɴ ᴜᴘ = .ᴍs–¹

(ɪɪ) ᴍᴀxɪᴍᴜᴍ ʜᴇɪɢʜᴛ ɪᴛ ʀᴇᴀᴄʜᴇs

ᴜsɪɴɢ, ᴀs = ᴠ² – ᴜ², ᴡᴇ ɢᴇᴛ

s = ᴠ²- ᴜ²/ᴀ

s = – . × ./(- × .)

s = . ᴍ

ᴛʜᴜs, ᴍᴀxɪᴍᴜᴍ ʜᴇɪɢʜᴛ ɪᴛ ʀᴇᴀᴄʜᴇs = . ᴍ.

(ɪɪɪ) ᴘᴏsɪᴛɪᴏɴ ᴀғᴛᴇʀ sᴇᴄᴏɴᴅs

ᴛ = s.

ɪɴ s, ᴛʜᴇ ʙᴀʟʟ ʀᴇᴀᴄʜᴇs ᴛʜᴇ ᴍᴀxɪᴍᴜᴍ ʜᴇɪɢʜᴛ ᴀɴᴅ ɪɴ s ɪᴛ ғᴀʟʟs ғʀᴏᴍ ᴛʜᴇ ᴛᴏᴘ.

ᴅɪsᴛᴀɴᴄᴇ ᴄᴏᴠᴇʀᴇᴅ ɪɴ s ғʀᴏᴍ ᴍᴀxɪᴍᴜᴍ ʜᴇɪɢʜᴛ,

s = ᴜᴛ + /ᴀᴛ ²

s = + / × . ×

s = . ᴍ