A tennis ball was tossed from a height of 1.25 m with an initial velocity of 18.

Question

A tennis ball was tossed from a height of 1.25 m with an initial velocity of 18.0 m/s at an angle 35.0 degree above the horizontal. The ball hits the ground at a horizontal distance of 32.1 m from the launch point. Use 10.0 N/kg for g. Find in m/s the magnitude of the velocity of the ball when it is at the top most position of its trajectory. Find in m/s^2 the magnitude of the acceleration of the ball when it is at the top most position of its trajectory. How long in seconds is the ball in flight? What is in m/s the magnitude of the velocity with which the ball hits the ground? What is in m the highest point relative to ground reached by the ball?

Explanation / Answer

a)

at the top most point vy = 0

it has only vx

speed = vx = vo*costheta = 18*cos35 = 14.74 m/s

b)

at the top most point the acceleration = a = g = 10 m/^2

c)

T = X/(vo*costheta)

T = 32.1/(18*cos35)

T = 2.18 s

d)

vy = VOY + ay*T

Vy = (18*Sin35)-(10*2.18)

Vy = 11.5 m/s

Vx = VO*costheta = 14.74 m/s

speed = v = sqrt(vx^2+vy^2) = 18.7 m/s

e)

H = yo + voy^2/2g

H = 1.25 + (18*sin35)^2/(2*10)

H = 6.58 m

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