Question

Marcus throws a ball into the air at a speed of 15m/s. How high does it go? How long doesit take to return to Marcus' hand?​

Answer 1

height=11.46m

time=3s

Step-by-step explanation:

Hello!

To solve this problem we must consider the following.

1. The maximum distance traveled upwards corresponds to the moment when the final speed is zero.

2. The ball thrown by marcus has the effect of a constant downward acceleration called gravity (g = 9.81m / s), therefore we can use the equation that describes the movement of a body when it has constant acceleration

`(Vf^(2)-Vo^2)/(2.g) =y`

where

Vf = final speed

=0M/S

Vo = Initial speed

=15M/S

g=gravity=-9.81m/s^2

Y = displacement

solving

`(0^(2)-(15)^2)/(2.(-9.81)) =y`

Y=11.46m

3.To calculate the time it takes for the ball to return to Marcus is the sum of the time it takes to go up plus the time it takes to go down.

for the time it takes for the ball to go up we use the following equation

Vf=Vo+a.t

solving for time

`T=(Vf-Vo)/(g)`

Vf=0m/S

Vo=15m/s

g=gravity=-9.81m/s^2

`T=(Vf-Vo)/(g)=(0-15)/(-9.81) =1.53s`

4.for the time it takes to go down, we must take into account that we already know the distance, the gravity and that the initial velocity is 0, we use the following equation

`y=VoT+0.5gt^(2)`

where

y=displacement=11.46M

Vo=0m/s

g=gravity=-9.81m/s^2

t=time

solving for time

`y=0.5gt^(2)\\t=\sqrt{(y)/(0.5g) } \\t=\sqrt{(11.46)/(0.5(9.81m/s^2)) \\`

t=1.53s

5. finally we add the time it takes for the ball to go up and down

T=1.53+1.53=3s