Question

A student throws a set of keys vertically upward to her sorority sister, who is in a window 4.00 m above. The keys are caught 1.50 s later by the sister's outstretched hand. (a) With what initial velocity were the keys thrown?(b) What was the velocity of the keys just before they were caught?

Answer 1

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).