A ball of mass m is thrown straight up into the air with an initial speed v_0. (

Question

A ball of mass m is thrown straight up into the air with an initial speed v_0. (a) Find an expression for the maximum height reached by the ball in terms of v_0 and g. (b) Using conservation of energy and the result of part (a), find the magnitude of the momentum of the ball at one-half its maximum height in terms of m and v_0. p = Drops of rain fall perpendicular to the roof of a parked car during a rainstorm. The drops strike the roof with a speed of 13 m/s, and the mass of rain per second striking the roof is 0.064 kg/s. (a) Assuming the drops come to rest after striking the roof, find the average force exerted by the rain on the roof. magnitude The response you submitted has the wrong sign N direction (b) If hailstones having the same mass as the raindrops fall on the roof at the same rate and with the same speed, how would the average force on the roof compare to that found in part (a)? (Assume the hailstones bounce back up off the roof.) The magnitude would be the same as part (a). The magnitude would be double that in part (a) The magnitude would be fourth times in part (a). The magnitude would be that in part (a).

Explanation / Answer

a )

hmax = Vo2 / 2 g

b )

using conservation of energy

0.5 m V2 = 0.5 m Vo2 - m g hmax / 2

0.5 m V2 = 0.5 m Vo2 - 0.25 m Vo2   

V2 = Vo2 - 0.5 Vo2

V = V0/ 21/2

P = m V

P = m Vo / 21/2

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a )

we have F = m a

F = 13 X 0.064

F = 0.832 kg.m/sec2

b )

Option "b" is correct

beacuse the force can be experienced by roof is double,

then the roof absorbs from hail particle is also double...

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