## I am struggling to answer these problems.?

1. A physcis student throws a ball straight up into the air with a speed of 17.5 m/s. The ball is in the air for a total of 3.60s before it is caught at its original position. How high does the ball rise?

2. A soccer ball moving with an initial speed of 1.5 m/s is kicked with a uniform acceleration of 6.3 m/s^2, so that the ball’s new speed is 9.2 m/s. How far has the soccer ball moved.

## Answers ( 4 )

During one half of this time, its velocity decreases from 17.5 m/s to 0 m/s. Since the ball’s deceleration is constant, let’s use the following equation to determine the maximum distance the ball rises.

d = ½ * (vi + vf) * t, vf = 0

d = ½ * 17.5 * 1.8 = 15.75 meters

Let’s use the following equation to determine the deceleration.

vf = vi + g * t

0 = 17.5 + g * 1.8

g = -17.5 ÷ 1.8

This is approximately -9.72 m/s^2. If you use the exact value of g in the following equation, you will get exactly the same distance.

d = vi * t + ½ * g * t^2

d = 17.5 * 1.8 + ½ * (-17.5 ÷ 1.8) * 1.8^2 = 15.75 meters

2. A soccer ball moving with an initial speed of 1.5 m/s is kicked with a uniform acceleration of 6.3 m/s^2, so that the ball’s new speed is 9.2 m/s. How far has the soccer ball moved.

Let’s use the following equation to determine the distance.

vf^2 = vi^2 + 2 * a * d

9.2^2 = 1.5^2 + 2 * 6.3 * d

12.6 * d = 82.39

d = 82.39 ÷ 12.6

This is approximately 6.54 meters.

1. A physcis student throws a ball straight up into the air with a speed of 17.5 m/s. The ball is in the air for a total of 3.60s before it is caught at its original position. How high does the ball rise?

h = Vy^2/2g = 17.5^2/19.6) = 15.63 m

check :

t/2 = √2h/g = √15.63*2/9.8 = 1.79 sec (rounded up to 1.8 sec)

2. A soccer ball moving with an initial speed of 1.5 m/s is kicked with a uniform acceleration of 6.3 m/s^2, so that the ball’s new speed is 9.2 m/s. How far has the soccer ball moved.

accel. time t = (9.2-1.5)/6.3 = 1.222 sec

distance d = Vo*t+a/2*t^2 = 1.5*1.222+3.15*1.222^2 = 6.54 m

The (physics) question is:

Bubba (m = 80 kg) is working construction on a third-floor scaffolding (h = 9 m). He wants to

get down to the ground floor. One quick way of getting down is to step off the scaffolding. Another

(maybe slower) way is to grab a nearby rope, which goes over a frictionless pulley on the 5th floor

and down to a barrel full of trash on the ground. The barrel full of trash has a mass mb = 60 kg,

so it will go up as Bubba comes down.

5. Bubba hit the ground just hard enough that he let go of the rope. The barrel full of trash

—remember the barrel? It was on the other end of the rope— is now directly above Bubba

and nothing is holding it up. How fast will the barrel be moving when it hits the ground?

(Hint: 13.3 m/s is incorrect.)

My answer is:

V2=U2+2ah

since he starts from rest ,initial velocity is U=0

V2=0+2*9.8*9

V=13.3 m/s

But the question clearly states that’s the wrong answer. What am I doing wrong and how do I fix it?

Perhaps the hint lies in the fact that he lets go of the rope and the barrel is still moving upward when he does find acceleration of man/barrel system as man descends tension is the same on both sides of pulley

T-man = T/barrel

m g – m a = m g + m a

80(9.81) – 80a = 60(9.81) + 60a

140a = 196.2

a = 1.40 m/s/s

find v of barrel at release

Vf^2 = Vi^2 + 2 a d

Vf^2 = 0 + 2(1.40)(9) = 25.226

Vf = 5.02 m/s

so the barrel is going up at 5.02 m/s when rope is released

find v of barrel back at ground

Vf^2 = 5.02^2 + 2(-9.81)(-9) = 201.8

Vf = 14.2 m/s

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