You and your friend are on the roof of a 20 m tall building investigating a free fall. Your friend releases a small rock from rest. You throw another rock simultaneously with your friend aiming it straight down.
(a) What initial speed should you supply to the rock so it hits the ground one second before the friend's one? Use g = 10 m/s^2.
time for first rock
-20 = -5 t^2
t = 2 seconds to fall
so t for second ball = 2 - 1 = 1 second
-20 = -Vi t + -5 t^2
20 = -Vi + 5
Vi = -15 m/s (the - sign means speed is down)
To determine the initial speed of the rock you throw, we can use the equations of motion for free fall. We need to consider the time of flight, the initial velocity, and the height of the building.
Given:
Height of the building (h) = 20 m
Acceleration due to gravity (g) = 10 m/s^2
Time it takes for your friend's rock to reach the ground (t) = 1 second
Using the equation of motion for falling objects:
h = ut + (1/2)gt^2,
where:
h is the height,
u is the initial velocity,
t is the time, and
g is the acceleration due to gravity.
For your friend's rock, we have:
20 = 0 + (1/2)(10)(1)^2,
20 = 5.
We can conclude that the time it takes for your friend's rock to reach the ground is 1 second, as it has already been given.
Now, for your rock to hit the ground one second before your friend's rock, the time of flight for your rock should be t - 1 = 1 - 1 = 0 second.
Using the same equation of motion:
h = ut + (1/2)gt^2.
For your rock:
20 = u(0) + (1/2)(10)(0)^2,
20 = 0.
This equation implies that your rock must be thrown downward with an initial velocity greater than 0 (since h = 20 and t = 0). The initial velocity should be such that your rock reaches the ground in exactly 0 seconds.
Therefore, to find the initial velocity needed for your rock to hit the ground one second before your friend's rock, you need to supply an initial velocity greater than 0, but less than that of your friend's rock.