Astudent throws a set of keys vertically upward to his fraternity brother, who is in a window 4.00 m above. the brother’s outstretched hand catches the keys 1.50 s later. (a) with what initial velocity were the keys thrown? (b) what was the velocity of the keys just before they were caught?

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

Ais the answer hope this