A 200g ball is launched from a heaight of 200m aboce a lake

A 200-g ball is launched from a heaight of 20.0m aboce a lake. Its launch angle is 40-degrees and it has an intial kineticenergy of 90.0-J. (a) Ues energy methods to determine its maximum height above the lake surface. (b) use projectile motion kinematics to repeat part. (c) Use energy methods to determine its speed just before impact with the water. (d) Repeat part (c) using projectile motion kinematics.

Solution

Initial kinetic energy = 0.5*m*v^2 = 90

Hence, v = sqrt[180/0.2] = 30 m/s

a.) The ball will undergo no acceleration towards the horizontal direction and its velocity along the vertical will be zero at the highest point.

Hence, the horizontal velocity of the ball at the highest point = 30 cos40 = 22.98 m/s

Equating the energy at points = 0.5 * 0.2 * 22.98^2 + 0.2*9.81*H = 90

Hence H = 18.95 above the launch point. That is height above lakes surface = 20 + 18.95 = 38.95 m

b.) Here, the vertical velocity = 30 sin40

The ball slows down to zero velocity along the vertical under action of g

Hence 0 = U^2 - 2*g*H

That is, H = 18.95 above launch point. Therefore, height above lake surface = 38.95 m

c.) Here again, we equate the energy at two points

90 + 0.2 * 9.81 * 20 = 0.5*0.2 * v^2

Hence, V = 35.95

d.) Here, the horizontal velocity remains same.

The vertical velocity will decrease and then increase in the downward direction. The total displacement before impact would be -20 m

Hence V^2 = U^2 + 2*g*20

V^2 = 371.86 + 392.4; Hence, V = 27.65

Therefore net speed = sqrt (27.65*27.65 + 22.98*22.98) = 35.95 m/s