Question

Charlie kicks a soccer ball up a small incline. On the way up, ball's acceleration has magnitude
jaj = 0:45 m/s2 and is directed in downhill direction. Charlie kicks the ball at the bottom of the
incline and then immediately start to walk up the incline with constant speed. Charlie performs
twi dierent trials.
a) In the first trial, Charlie kicks the ball with initial speed v0 = 3:4 m/s. Charlie is 2.3-m behind
the ball when the ball is at the highest point. What is the speed vC of Charlie?
b) Charlie performs the second trial. He kicks the ball with unknown speed v00
but walks with the same speed vC as in the first trial. Charlie is now 0.8 m behind the ball when the ball is at the
highest point. What is the initial speed v0 of the ball at the bottom of the hill? (Hint: You need to set-up a quadratic equation for v0).

Answer 1

If acceleration of the soccer ball at slope is 0.45 m/s² , then angle of inclined surface is calculated as given below

g sin = 0.45 m s^-2 ,   hence = sin^-1 (0.45 / 9.8 ) = 2.6^o

(a) First trial

Kicking speed = 3.4 m/s ,

Distance travelled on the inclined plane is obtained from equation of motion " v² = u² - 2a S " , where v is final speed which is zero when it reaches the maximum distance , u is initial speed , a is retardation and S is distance travelled

S = u² / ( 2 a ) = ( 3.4 3.4 ) / ( 2 0.45 ) = 12.84 m

Time taken by the ball to travel maximum distance is obtained from equation of motion " v = u - (a t ) "

If final speed v = 0, then time t = u/a = 3.4/0.45 = 7.56 s

If Charlie is 2.3 m behind the ball when the ball reaches highest point ,

then Charlie has walked a distance ( 12.84 - 2.3 ) m = 10.54 m in 7.56 s

Hence walking speed of Charlie = 10.54 / 7.56 1.4 m/s

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Trial(2)

Let u is kicking speed of ball

If t is time to reach maximum height , distance S travelled by the ball on inclined plane is given by

S = [ u t ] - [ (1/2) a t² ]

During this time t , charlie walked with speed 1.4 m/s and he is 0.8 m behind the ball when the ball reaches maximum height

Hence we get, [ u t ] - [ (1/2) a t² ] = (1.4 t + 0.8 ) .............(1)

from equation of motion, " v = u - at " , final speed v = 0 when ball reaches maximum height ,

we have u = at

Hence by substituting u = a t , in equation (1) , we get

(1/2) a t² = 0.225 t² = (1.4 t + 0.8 )

Hence we geta quadratic equation , t² - 6.222 t - 3.56 = 0

we get t = 6.75 s by solving above quadratic equation

Hence initial kicking speed u = a t = 0.45 6.75 = 3.04 m/s