Question

A ball us thrown vertically upwards with a velocity of 20 m/s from the top of a multi-storey building. The height of the point from where the ball is thrown is 25 m from the ground. (1) how high the ball will rise? (2) How long it will be before the ball hits the ground?

Answer 1

Intial Velocity (u)=20 m/s

Final Velocity at highest point it reach(v)=0 m/s²

Accerlation due to gravity=-10 m/s²

Here We take Accerlation as -10 as while going upward it Velocity will Decrease

Use Equation of Motion

v²=u²+2as

s=height

0²=20²+2×-10×s

0²-20²=-20s

0-400=-20s

s=-400/-20

s=20 m

So, from Multi-storey building ball reach=20 m.

And from Ground ball is at height of=20+25=45 m.

(2) While Returning back on Earth

Intial Velocity (u)=0 m/s²

It have to Hit ground means It have to go 45 m down.

s=45 m

Accerlation due to gravity (a)=10 m/s²

Time taken to Reach ground=t

Use Equation of motion

$s = ut + \frac{1}{2} {at}^{2} \\ \\ 45 = 0 \times t + \frac{1}{2} \times 10t {}^{2} \\ \\ 45 = {5t}^{2} \\ \\ {t}^{2} = \frac{45}{5} \\ \\ {t}^{2} = 9 \\ \\ t = \sqrt{9} \\ \\ t = 3$

Time=3 seconds

Answer 2

Answer:

t=5

Explanation:

To supplement the previous answer, the user forgot to include the amount of time taken to throw the ball upward as well, which would add to the total time before the ball hits the ground.

To calculate the time that the ball is in the air, we use the equation

V(final)=V(initial) +at

- V(final) = 0, because that is when the ball has reached maximum height and starts to fall back to the ground

- V(initial) = 20, as given in the description

- a, (acceleration of gravity) = -10m/s^2

Substitute the given values

0=20 + (-10)t

Solve for t

t= -20/-10 = 2

t=2 (This number represents the time that the ball takes after being thrown into the air to come to a stop and then fall back to Earth)

Taking the value of the freefall as calculated in the previous answer (t=3), and adding it to the answer just calculated (t=2), we receive a total in air time of 5 seconds for the ball.