Question

A ball is thrown vertically upwards from the top of a tower at 4.9 ms1. It strikes the pond near the base of the tower after 3 seconds. The height of the tower is [take g = 9.8 m/s2]

Answer 1

Answer:

Height of the tower = 29.4 meter

Step by step explanations :

Given that,

A ball is thrown vertically upwards from the top of a tower at 4.9 m/s

here,

initial velocity of the ball = 4.9 m/s

let the time taken by the ball to reach its maximum height be t

so,

at maximum height it's velocity will 0

here,

we have

Initial velocity(u) = 4.9 m/s

final velocity(v) = 0 m/s

gravitational acceleration(g) = -10 m/s²

we know that,

time taken by object to reach maximum height = time taken to reach its initial position

so,

time taken by ball to pass its initial position = 2 × time taken to reach maximum height

now,

by the equation of motion,

v = u + gt

putting the values,

0 = 4.9 + (-9.8)t

-9.8t = -4.9

t = -4.9/-9.8

t = 0.5 s

now time taken to pass its initial position = 0.5 × 2

= 1 s

now given the total time taken by ball to reach ground = 3 s

and we know that,

velocity by which any object when it throws vertically upwards = velocity at which it passes it initial position

so,

from the initial position

initial velocity of ball = 4.9 m/s

time taken to reach ground = 3 - 1

= 2 s

gravitational acceleration(g) = 10 m/s²

let the height of the tower be h

by the gravitational equation of motion,

h = ut + ½ gt²

again putting the values,

h = 4.9(2) + ½ × 9.8 × 2 × 2

h = 9.8 + 19.6

h = 29.4 m

so,

Height of the tower = 29.4 meter

Answer 2

Answers ➡29.4 m

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Hᴇʀᴇ, Yᴏᴜʀs Sɪᴍᴘʟᴇ & Cᴀᴛᴄʜʏ Sᴏʟᴜᴛɪᴏɴs—

Given & to find-

Ininitial speed(u) = - 4.9 m/s. (upward hence taken as - ive)

height of tower (s) = ?

Time(t) = 3 sec

g = 9.8 m/s²

Using laws of motion

S = ut + 1/2 gt²

Putting the value

S = - 4.9×3 + 1/2 ×9.8 ×3²

On solving

S = 29.4 m

Hence, the height of tower is 29.4 m