Question

A ball is dropped from a tower of height 30.0 m. If the terminal speed of the ball is 25.0
m/s then calculate the time taken by the ball to hit the ground.
(a) 2.67 s
(b) 3.58 s
(c) 2.47 s
(d) 5.54 s​

Answer 1

Answer:

a

Explanation:

s=ut + 1/2 at^2

a= 10m/s^2

30=0 * t + 1/2 *10 *t^2

30 = 5t^2

t^2= 30/5

t = root over 6

t = 2.67 (approx)

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Answer 2

Answer:

c 2.47

Explanation:

PART 1

Terminal speed is the maximum speed of a body or the final speed of body.

Here, we will need to find time taken by body to attain terminal speed,

v=u+at

25=0+10t

therefore,t=2.5 sec

Now, Height lost in 2.5 sec;

S=ut+1/2at^2

S=1/2*10*2.5*2.5

S=31.25

PART 2

Here Height lost is grater than height of tower so, it will hit ground before attaining terminal velocity.

hence,

by using S=ut+1/2at^2;

30=1/2*10*t^2

hence

t^2=6

t=2.47

only one option is less than 2.5 so without solving part 2  we get ans.