A baseball is thrown vertically upward with an initial velocity of 20m/s. The maximum height gained by the ball is?
I do not understand this problem!
To solve this problem, we can use the kinematic equation for vertical motion:
v^2 = u^2 + 2as
where:
v = final velocity (when the ball reaches its highest point, the velocity is 0 m/s)
u = initial velocity (20 m/s)
a = acceleration (acceleration due to gravity, which is -9.8 m/s^2 because it is acting in the opposite direction of the ball's motion)
s = displacement (maximum height gained by the ball)
We can rearrange the equation to solve for s:
s = (v^2 - u^2) / (2a)
Plugging in the values:
s = (0^2 - 20^2) / (2 * -9.8)
s = (-400) / (-19.6)
s = 20.41 meters
Therefore, the maximum height gained by the ball is approximately 20.41 meters.
If you have learned about energy, you can calculate by equating kinetic and potential energies assuming all kinetic energy (due to velocity) is converted to potential energy (due to height).
mgh=(1/2)mv²
when m cancels out,
g=acc. due to gravity (m/s/s)
h=height attained (m)
v=initial velocity
final velocity is zero at the highest point.
h=(1/2)v²/g