a) 10.0 m/s
b) -4.7 m/s
The formula for distance under constant acceleration is
d = 0.5AT^2
The formula for distance with a specified velocity is
d = VT
So the distance the keys travel with an initial velocity and under constant acceleration by gravity is
d = VT - 0.5AT^2
The acceleration due to gravity is 9.8 m/s^2 and the time T is 1.50 s, and finally, the distance traveled is 4.00 m. So substitute those values into the equation and solve for V
d = VT - 0.5AT^2
4.00m = 1.50s * V - 0.5 * 9.8 m/s^2 * (1.5s)^2
Do the multiplications
4.00m = 1.50s * V - 4.9m/s^2 * 2.25 s^2
Cancel the s^2 terms
4.00m = 1.50s * V - 4.9m * 2.25
Do the multiplication
4.00m = 1.50s * V - 11.025m
Add 11.025m to both sides
15.025m = 1.50s * V
Divide both sides by 1.50s
10.01667 m/s = V
Since we have 3 significant figures in the data, round results to 3 significant figures.
V = 10.0 m/s
So the keys were initially thrown upwards with a velocity of
10.0 m/s
Since it took 1.50 seconds from launch to catch, the velocity of the keys will decrease by 9.8 m/s^2 times the time. So
V = 10.0 m/s - 1.50s * 9.8 m/s^2
V = 10.0 m/s - 14.7 m/s
V = -4.7 m/s
So at the time the keys were caught, they were moving downward at a velocity of 4.7 m/s
