A tennis ball is shot vertically upward from the surface of an atmosphere-free planet with an initial speed of 20.0 m/s. One second later, the ball has an instantaneous velocity in the upward direction of 15.0 m/s. What is the magnitude of the acceleration due to gravity on the surface of this planet.
4.06
To find the magnitude of the acceleration due to gravity on the surface of the planet, we can use the information provided.
Given:
Initial velocity, u = 20.0 m/s
Velocity after 1 second, v = 15.0 m/s
We can use the equation of motion:
v = u + at
where:
v = final velocity
u = initial velocity
a = acceleration
t = time
Rearranging the equation, we have:
a = (v - u) / t
Substituting the given values:
a = (15.0 m/s - 20.0 m/s) / 1 s
a = -5.0 m/s / 1 s
Therefore, the magnitude of the acceleration due to gravity on the surface of this planet is 5.0 m/s². Note that the negative sign indicates that the acceleration is in the opposite direction of the initial velocity.
To solve this problem, we need to use the equations of motion for vertical motion.
The equation we will use is:
v = u + at
Where:
v = final velocity
u = initial velocity
a = acceleration
t = time
According to the problem, the ball has an initial velocity (u) of 20.0 m/s and a final velocity (v) of 15.0 m/s.
However, we need to find the acceleration due to gravity (a) on the planet's surface. Since the ball is moving upwards, the acceleration due to gravity will be acting in the opposite direction, downwards.
So, we can rewrite the equation as:
v = u - at
Rearranging the equation to solve for acceleration (a):
a = (u - v) / t
Plugging in the given values, we have:
a = (20.0 m/s - 15.0 m/s) / 1 s
a = 5.0 m/s²
Therefore, the magnitude of the acceleration due to gravity on the surface of this planet is 5.0 m/s².