A rock is thrown straight up into the air with an initial speed of 53 m/s at time
t = 0.
Ignore air resistance in this problem. At what times does it move with a speed of 40 m/s? Note: There are two answers to this problem.
earlier time:
later time:
v = 53-9.8t
so find when v = ±40
To find the times at which the rock moves with a speed of 40 m/s, we can use the kinematic equation for velocity:
v = u + at
Where:
v = final velocity (40 m/s)
u = initial velocity (53 m/s)
a = acceleration (due to gravity, -9.8 m/s^2)
t = time
First, let's find the time when the rock reaches its highest point (when its vertical velocity becomes 0).
Using the equation:
v = u + at
0 = 53 - 9.8t
Rearranging the equation, we have:
9.8t = 53
t = 53 / 9.8
t ≈ 5.41 seconds
This is the time it takes for the rock to reach its peak.
Now let's find the earlier and later times when the rock has a speed of 40 m/s.
For the earlier time:
v = u + at
40 = 53 - 9.8t
9.8t = 53 - 40
9.8t = 13
t = 13 / 9.8
t ≈ 1.33 seconds
For the later time:
v = u + at
40 = 53 - 9.8t
9.8t = 53 - 40
9.8t = 13
t = 13 / 9.8
t ≈ 1.33 seconds
Therefore, the earlier time is approximately t = 1.33 seconds and the later time is also approximately t = 1.33 seconds.