In the figure here, a ball is thrown up onto a roof, landing 4.80 s later at hei

Question

In the figure here, a ball is thrown up onto a roof, landing 4.80 s later at height h = 19.0 m above the release level. The ball's path just before landing is angled at = 53.0 with the roof. (a) Find the horizontal distance d it travels. (Hint: One way is to reverse the motion, as if it is on a video.) What are the (b) magnitude and (c) angle (relative to the horizontal) of the ball's initial velocity?

This answer has no unitsdegreesmkgsm/sm/s^2NJWN/mkg•m/sPakg/m^3gm/s^3times

This answer has no unitsdegreesmkgsm/sm/s^2NJWN/mkg•m/sPakg/m^3gm/s^3times

This answer has no unitsdegreesmkgsm/sm/s^2NJWN/mkg•m/sPakg/m^3gm/s^3times

Explanation / Answer

Let utilise the hint given. Let us reverse the motion,as if the ball was released from the roof at an angle of 53o and total time of travel is 4.8 seconds. The roof height is 19m.

a) The the velocity at which the ball is thrown from the roof be v

time taken by ball to reach highest point =t= vsin53 / g = 0.8v/g

Maximum height(measured from roof) reached by ball = (vsin53)2/2g = 0.64v2/2g

Maximum height reached by ball measured from ground = 19+0.64v2/2g

Let, Time taken by ball to reach ground from heighest point = t1

19+0.64v2/2g = (1/2)gt12

t1 = sqrt[(38+0.64v2)/g]

t+t1 = 4.8

Or, 0.8v/g + sqrt[(38+0.64v2)/g] = 4.8

Or, sqrt[(38+0.64v2)/g)] = 4.8-0.8v/g

Or, (38+0.64v2)/g = (4.8-0.8v/g)2 = 23.04-7.68v/g+0.64v2/g2

g=9.8m/s2

putting this value in above equation we get

3.877+0.065v2 = 23.04-0.784v+0.00666v2

or, 0.05834v2+0.784v-19.163=0

v = 12.6m/s

Horizontal distance d travelled = 12.6cos53 * 4.8 = 36.288m

b)t1 = sqrt[(38+0.64v2)/g] = sqrt[(38+0.64*12.62)/9.8] = 3.77 seconds

vertical component of velocity at ground = 9.8*3.77 = 36.95 m/s(upwards)

horizontal component of velocity at ground = 12.6cos53 = 7.56 m/s(towards right)

Magnitude of velocity = sqrt(36.952+7.562) = 37.71 m/s

c)angle of ball at initial position = tan-1(36.95/7.56) = 78.4o

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