A ball is shot upwards with an initial speed of 31 m/s. While it is in the air it experiences a constant downwards acceleration of 10 m/s2. Find its velocity 1, 2, 3, and 4 seconds after being released.
To find the velocity of the ball at different time intervals, we can use the equation of motion known as the kinematic equation:
v = u + at
where,
v = final velocity
u = initial velocity
a = acceleration
t = time interval
Given:
u = 31 m/s (initial velocity)
a = -10 m/s^2 (constant downwards acceleration)
To find the velocity at different time intervals, we substitute the given values into the equation and solve for v.
1. Velocity after 1 second:
t = 1 second
v = u + at = 31 + (-10)(1)
v = 31 - 10
v = 21 m/s
2. Velocity after 2 seconds:
t = 2 seconds
v = u + at = 31 + (-10)(2)
v = 31 - 20
v = 11 m/s
3. Velocity after 3 seconds:
t = 3 seconds
v = u + at = 31 + (-10)(3)
v = 31 - 30
v = 1 m/s
4. Velocity after 4 seconds:
t = 4 seconds
v = u + at = 31 + (-10)(4)
v = 31 - 40
v = -9 m/s
Therefore, the velocity of the ball at 1, 2, 3, and 4 seconds after being released is 21 m/s, 11 m/s, 1 m/s, and -9 m/s, respectively.