A football player throws a 400 g football at 30 degree C to the horizontal. Bein

Question

A football player throws a 400 g football at 30 degree C to the horizontal. Being in a god physical condition, he is able to develop 200 Watts of power for the 0.6 s that it takes him to throw the half. The ball is caught at exactly the same height above the ground as it was thrown. How far does the ball travel in the horizontal direction? i) What amount of initial energy does the ball receive from the athlete? Ii) What is the initial potential of the ball (immediately after it has separated from the athlete's hand)? Iii) What is the initial kinetic energy of the ball (immediately after it has separated from athlete's hand?) iv) What is the initial speed of the ball (immediately after it has separated from the athlete's hand)? v) What is the vertical component of the ball's velocity? Vi) How long will it take for the ball whose initial velocity was found in v) to get to the maximum height? Vii) How long will it take for the ball to get up to the maximum and then fall back to the initial height? Viii) What is the horizontal component of the ball's initial velocity? Ix) How far will be ball travel in the horizontal direction within the time found in viii)?

Explanation / Answer

P = KE/t

KE = P*t = 200*0.6 = 120 J

KE = (1/2)*M*V^2 = 120

V = sqrt(2*120)/0.4)

V = 24.5 m/s

as the ball is reaching the same level

vertical displacement = 0

dy = voy*T - 0.5*g*T^2

T = 2Voy/g

horizantal distance travelled during this time T

ax = o

X = Vox*T

X = Vox*2*Voy/g

X = V*cos30*V*sin30*2/g

x = v^2*sin60/g

X = 5.302 m

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