Question

A ball of mass m is dropped from the top of tower of suitable height, the ratio of distance covered by the ball in first 3 s to the next 3 s is

Answer 1

HERE IS YOUR ANS -
distance covered in first 3 seconds-
intial speed = 0

✨by second equation of straight line motion-

S = Ut+1/2at². (g = 10)
s= 0t + 1/2g(3)²
= 1/2×10×9 = 45

distance covered by ball in next 3 seconds-

by 2nd equation of straight line motion-

s = Ut+1/2at²
= 30(3)+1/2(10)(3)²
s = 135

so ,,ratio = 45/135 = 1:3 ANSWER...

HOPE IT HELPS YOU ✨...

Answer 2

Answer:

The ratio of distance covered by the ball in the first 3 s to the next 3 s is 1:4.

Explanation:

The distance traveled by the body is given as,

$S=ut+\frac{1}{2}gt^2$       (1)

Where,

S=distance covered by the ball

u=initial velocity of the ball

t=time taken by the ball to drop

g=acceleration due to gravity=10m/s²

From the question we have,

u=0        

t₁=3 second

t₂=3+3=6second

For t₁=3 second

By placing the values in equation (1) we get;

$S_1=0\times 3+\frac{1}{2}\times 10\times (3)^2$

$S_1=\frac{1}{2}\times 10\times 9$    

$S_1=45m$     (2)

For t₂=6 second

By placing the values in equation (1) we get;

$S_2=0\times 6+\frac{1}{2}\times 10\times (6)^2$

$S_2=\frac{1}{2}\times 10\times 36$

$S_2=180m$       (3)

Now the ratio of equations (2) and (3) we get;

$\frac{S_1}{S_2}=\frac{45}{180}$

$\frac{S_1}{S_2}=\frac{1}{4}$        (4)

Hence, the ratio of distance covered by the ball in the first 3 s to the next 3 s is 1:4.

#SPJ2