A golf ball with a weight of 01lbf is launched at a 30 degre

A golf ball with a weight of 0.1lbf is launched at a 30 degree angle as shown. Show that the initial velocity of the golf ball is v_o136ft/s.

Assume that the golf club is in contact with the ball for 0.0005 seconds. Assume that the impulse force is constant. Using impulse-momentum, find the impulse force. Ignore the impulse due to the weight of the ball. Ans. F850lbf.

A baseball with a mass of 0.15kg is thrown to the batter at an initial velocity of v₁=30m/s (to the left and at a downward angle of 15 degrees). The batter hits the ball when it is at an initial height of y_o=0.75m. It launches from this point at an upward angle of 30 degrees and a velocity v₂. It travels horizontally a distance of 100m, and is caught there at a height of y=2.5m. Show that v₂34m/s.

Assume that the bat was in contact with the ball for 0.00075 seconds. Assume that the impulse force was constant during this time. Using impulse-momentum, find the impulse force is the x & y directions and the net impulse force. Ignore the impulse due to the weight. Ans. Is Fx=11.7kN, Fy=4.97kN, F=12.7kN.

Solution

Calculate initial velocity of golf ball

R = Vo ^2 sin2theta / g

We know R = 500 theta = 30° and g=32.174 ft/s2

Therefore applying in above formula we get

500 = Vo ^2 sin (2*30) / 32.174

Vo ^2 = 18575.6 675

Vo = 136.2925 ft/s

Now for velocity of base ball

R = V2 ^2 sin 2 theta /g

We have R=100 and theta 30° take g= 9.81m/s2

Applying we get

V2^2 =1132.76122

And V2 = 33.6565 m /s nearly equal to 34m/s