A 2.25 kg ball experiences a net force of 965 N up a ramp as shown. Once the ball reaches the top of the ramp, the force no longer acts. The force acts over a distance of 1.50 m on the ramp. Find the horizontal distance x that the ball travels before it hits the deck. The top of the ramp is 4.50 meters above the deck below.

consider the motion of ball on the ramp ::

a = net acceleration on ramp = Net force / mass = 965/2.25 = 428.89 m/s2

d = length of the ramp = 1.50 m

Vi = initial velocity = 0

Vf = final velocity at the top of the ramp

Using the equation

Vf2 = Vi2 + 2 a d

Vf2 = 02 + 2 (428.89) (1.50)

Vf = 35.87 m/s

consider the motion of ball after it leaves the ramp ::

Vix = Vf Cos28 = component of velocity in X-direction = 35.87 Cos28 = 31.67

Viy = Vf Sin28 = component of velocity in Y-direction = 35.87 Sin28 = 16.84

consider the vertical motion of the ball:

Y = vertical displacement = height of top of ramp above the deck = - 4.50 m

a = - 9.8

Using the equation

Y = Viy t + (0.5) a t2

- 4.50 = 16.84 t + (0.5) (-9.8) t2

t = 3.686

distance travelled in X-direction is given as ::

X = Vix t = 31.67 x 3.686 = 116.74 m