Question

A student throws a set of keys vertically upward to her sorority sister, who is in a window 3.50 m above. The second student catches the keys 1.10 s later. (a) With what initial velocity were the keys thrown? magnitude m/s direction (b) What was the velocity of the keys just before they were caught? magnitude m/s direction

Answer 1

a) 8.58 m/s upward

b) -2.211 m/s downward

Step-by-step explanation:

Let gravitational acceleration g = -9.81m/s2. This is negative because it deceleration the upward motion of the key.

(a)We have the following equation of motion

`s = v_0t + gt^2/2`

where
`v_0` is the initial upward velocity of the keys, t = 1.1s is the time it takes for the keys to travel a distance of s = 3.5 m

`3.5 = v_01.1 - 9.81*1.1^2/2`

`3.5 = 1.1v_0 - 5.94`

`1.1v_0 = 3.5 + 5.94 = 9.44`

`v_0 = 9.44 / 1.1 = 8.58 m/s`

So the keys were thrown initially upward with a speed of 8.58 m/s

(b) If the initial velocity of the key is 8.58 m/s and it is subjected to a deceleration of 9.81m/s2 for 1.1s then the velocity right at the 1.1s instant is

`v = v_0 + gt = 8.58 - 9.81*1.1 = -2.211 m/s`

So they keys would have a downward speed of 2.211 m/s