Question

A stone dropped from the top of a tower reaches the ground in 8 (a= 10ms-2). calculate:। the height of the tower. (b). the distance travelled by the stone in the8 second​

Answer 1

Given :

[ A stone dropped from the top of a tower reaches the ground in 8 seconds ( taking acceleration due to gravity 10 ms⁻² ) ]

• initial velocity of stone, u = 0
• time taken by stone to reach the ground, t = 8 s
• acceleration due to gravity, a = 10 ms⁻²

To find :

• height of the tower = distance covered by stone in 8 sec = s

Formula required :

• Second equation of motion

     s = u t + 1/2 a t²

[ where s is distance covered, u is initial velocity, t is time taken, a is acceleration ]

Calculation :

Using second equation of motion

→ s = u t + 1/2 a t²

→ s = ( 0 ) ( 8 ) + 1/2 ( 10 ) ( 8 )²

→ s = 5 × 64

→ s = 320 m

Therefore,

height of tower is 320 m and also the distance covered by stone in 8 seconds is 320 m.

Answer 2

Answer:

answer is 320 m

Explanation:

given

initial velocity= 0

time=8sec

a=10ms-2

formula required

s=ut+1/2at2

put the value

s=(0)(8)+1/2(10)(8)2

S=(5)(64)

S=320 m