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

The purpose of a lab is to do your "expirements" the importance of the topic is so you can now what the story is about and the last one is because no one is me with my question ( what is 3 over 4 x - 9 = 27 maybe you can )