Question

A 2.25 kg ball expereinces a net force of 965 N up a ramp (images show 280 degree ramp). Once the ball reaches the top of the ramp, the force no longer acts. The force acts over the distannce 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.

Answer 1

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

V_i = initial velocity = 0

V_f = final velocity at the top of the ramp

Using the equation

V_f² = V_i² + 2 a d

V_f² = 0² + 2 (428.89)(1.50)

V_f = 35.87 m/s

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

V_ix = V_f Cos28 = component of velocity in X-direction = 35.87 Cos28 = 31.67

V_iy = V_f 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 = V_iy 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 = V_ix t = 31.67 x 3.686 = 116.74 m