1. A ball is thrown upwards at 20 m/s. If air resistance is negligible, how high

Question

1. A ball is thrown upwards at 20 m/s. If air resistance is negligible, how high does it rise?

2. A ball is dropped from a roof 6 m high. If air resistance is negligible, how fast is the ball going right before it hits the ground?   

3. A 90 kg climber climbs 45 m up a vertical wall.

   a. How much potential energy does the climber now have.

   b. If the climber continues climbing to a height of 85 m, how much potential energy does he now have?

4. The diagram below shows a 10,000 kg bus traveling on a straight road which rises and falls. The speed of the bus at point A is 26.82 m/s (60 mph). The engine has been disengaged and the bus is coasting. Friction and air resistance are assumed negligible. The numbers on the left show the altitude above sea level in meters. The letters A-F correspond to points on the road at these altitudes.

Find the speed of the bus at points B-F.

A ‘bad guy’ has planted a bomb on the bus. If the speed of the bus falls below 22.35 m/s (50 mph) the bomb will explode. Will the speed of the bus fall below this value and explode? If you feel the bus will explode, identify the interval (i.e B-C, D-E) in which the explosion occurs.

Explanation / Answer

1) Use conservation of energy

final potentail energy = initial kinetic energy

m*g*h = (1/2)*m*v^2

h = v^2/(2*g)

= 20^2/(2*9.8)

= 20.4 m

2) Use conservation of energy

Initial potentail energy = final kinetic energy

m*g*h = (1/2)*m*v^2

v = sqrt(2*g*h)

= sqrt(2*9.8*6)

= 10.84 m/s

3)

a) U = m*g*h1 = 90*9.8*45 = 39690 J

b) U = m*g*h2 = 90*9.8*85 = 74970 J

4)

VB = sqrt(VA^2 + 2*g*(hA - hB))

= sqrt(26.82^2 + 2*9.8*5)

= 28.59 m/s

vC = VA = 26.82 m/s

VD = sqrt(VA^2 + 2*g*(hA - hD))

= sqrt(26.82^2 - 2*9.8*5)

= 24.93 m/s

VE = sqrt(VA^2 + 2*g*(hA - hE))

= sqrt(26.82^2 - 2*9.8*10)

= 22.88 m/s

VF = CD = 24.93 m/s

The bomb will not explode.

because 22.88 > 22.35 m/s

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