Suppose that the gravitational acceleration on a certain planet is only 3.1 m/s2. A space explorer standing on this planet throws a ball straight upward with an initial velocity of 25 m/s.
(a) What is the velocity of the ball 4 seconds after it is thrown?
m/s
(b) How much time elapses before the ball reaches the high point in its flight?
s
a.37.4m/s b.5 s
How do you figure it out?
A.) v=vo+at
vo=25m/s. a=3.1 m/s^2 t=4s
v=(25m/s)+(3.1m/s^2)(4s)
v=(25m/s)+(12.4m/s)
v=25m/s+12.4m/s
v=37.4 m/s
To calculate the velocity of the ball 4 seconds after it is thrown, you can use the equation of motion:
v = u + at
Where:
v = final velocity
u = initial velocity
a = acceleration
t = time
Given:
u = 25 m/s (initial velocity)
a = -3.1 m/s^2 (acceleration, negative since it is acting opposite to the direction of motion)
t = 4 s (time)
(a) Substituting the given values into the equation:
v = 25 m/s + (-3.1 m/s^2)(4 s)
Calculate the value:
v = 25 m/s - 12.4 m/s
v = 12.6 m/s
Therefore, the velocity of the ball 4 seconds after it is thrown is 12.6 m/s.
To determine how much time elapses before the ball reaches the high point in its flight, you can use the equation for time:
t = (v - u) / a
Given:
u = 25 m/s (initial velocity)
v = 0 m/s (final velocity at the highest point)
a = -3.1 m/s^2 (acceleration, negative since it is acting opposite to the direction of motion)
(b) Substituting the given values into the equation:
t = (0 m/s - 25 m/s) / (-3.1 m/s^2)
Calculate the value:
t = -25 m/s / -3.1 m/s^2
t ≈ 8.06 s
Therefore, it takes approximately 8.06 seconds for the ball to reach the high point in its flight.