A man who is standing on the roof of a building throws a baseball vertically downward toward the sidewalk at a speed of 6.85m/s. The sidewalk is 34.9m below the roof. Calculate the speed of the ball just as it hits the pavement.
finalenergy=initial PE+intialKE
1/2 m v^2=mg*34.9+1/2 m 6.85^2
or v= sqrt (2*9.8*34.9+6.85^2)
To calculate the speed of the ball just as it hits the pavement, you can use the following formula:
v^2 = u^2 + 2as
Where:
- v is the final velocity (the speed of the ball just as it hits the pavement).
- u is the initial velocity (the speed at which the ball was thrown downward from the roof).
- a is the acceleration due to gravity (approximated as -9.8 m/s^2, where the negative sign indicates the downward direction).
- s is the displacement (the distance the ball travels from the roof to the sidewalk).
In this case, the initial velocity (u) is given as 6.85 m/s, the acceleration due to gravity (a) is -9.8 m/s^2, and the displacement (s) is 34.9 m (since the sidewalk is 34.9 m below the roof).
Now, substitute these values into the equation:
v^2 = (6.85)^2 + 2(-9.8)(34.9)
Calculating further:
v^2 = 46.9225 + (-682.04)
v^2 = -635.1175
Since velocity cannot be negative in this context, we can disregard the negative sign. Therefore, the magnitude of the velocity is:
v = √635.1175
v ≈ 25.217 m/s
So, the speed of the ball just as it hits the pavement is approximately 25.217 m/s.