You are on a bridge which is a height 30m above the water. You first drop a ball

Question

You are on a bridge which is a height 30m above the water. You first drop a ball and then 1.5 seconds later throw a second ball down so that both balls hit the water at the same time. (assume the positive y-direction is up) Part A At what time after you release the first ball does it hit the water? Part B With what velocity must the second ball be thrown downwards? Part C What is the speed of the first ball just before it hits the water? Part D What is the velocity of the second ball just before it hits the water?

Explanation / Answer

A)
d = Vi t + a t² / 2
where
d = distance = -30 m (negative because falling down)
Vi = initial velocity = 0 m/s
t = time = ?
a = acceleration by gravity = -9.8 m/s²
so
-30 = 0 t + (-9.8) t² / 2
-30 = -4.9 t²
t² = 6.122
t = 2.47 s

B)
d = Vi t + a t² / 2
where
d = distance = -30 m (negative because falling down)
Vi = initial velocity = ?
t = time = 2.47 - 1.5 = 0.97 m/s
a = acceleration by gravity = -9.8 m/s²
so
-30 = Vi 0.97 + (-9.8) 0.97² / 2
-30 = Vi 0.97 - 4.61
Vi = -35.68 m/s

C)
(Vf)² = (Vi)² + 2 a d
where
Vf = final velocity = ?
Vi = initial velocity = 0 m/s
a = acceleration by gravity = -9.8 m/s²
d = distance = -30 m (negative because falling down)
so
(Vf)² = 0² + 2 (-9.8) (-30)
(Vf)² = 588
Vf = -24.25 m/s

(the ball is falling down, so you must take the negative square root)
BUT, be careful... they are asking for the SPEED, not the velocity!

speed = velocity without indication of direction, so

Vf = 24,25 m/s

D)

(Vf)² = (Vi)² + 2 a d

where
Vf = final velocity = ?
Vi = initial velocity = -13.2 m/s
a = acceleration by gravity = -9.8 m/s²
d = distance = -30 m (negative because falling down)
so

(Vf)² = (-35.68)² + 2 (-9.8) (-30)
(Vf)² = 1861.06

Vf = 43.16 m/s

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