Answer:
a)The keys were thrown with an initial velocity of 10.0 m/s.
b)The velocity of the keys just before they were caught is -4.72 m/s (the keys were caught on their way down).
Explanation:
Hi there!
The equations for the height and velocity of the keys are the following:
h = h0 + v0 · t + 1/2 · g · t²
v = v0 + g · t
Where:
h =height of the keys after a time t.
h0 = initial height.
v0 = initial velocity.
t = time.
g = acceleration due to gravity (-9.81 m/s² considering the upward direction as positive)
v = velocity of the keys at time t.
a) We know that at t = 1.50 s, h = 4.00 m and let´s consider that the origin of the frame of reference is located at the point where the keys are thrown so that h0 = 0. Then, using the equation of height, we can obtain the initial velocity.
h = h0 + v0 · t + 1/2 · g · t²
4.00 m = v0 · 1.50 s - 1/2 · 9.81 m/s² · (1.50 s)²
4.00 m + 1/2 · 9.81 m/s² · (1.50 s)² = v0 · 1.50 s
v0 = ( 4.00 m + 1/2 · 9.81 m/s² · (1.50 s)²) / 1.50 s
v0 = 10.0 m/s
The keys were thrown with an initial velocity of 10.0 m/s.
b) Now, using the equation of velocity we can calculate the velocity at t = 1.50 s:
v = v0 + g · t
v = 10.0 m/s - 9.81 m/s² · 1.50 s
v = -4.72 m/s
The velocity of the keys just before they were caught is -4.72 m/s (the keys were caught on their way down).
