A ball is launched with an initial velocity of 26.4 m/s at a 45° angle from the

Question

A ball is launched with an initial velocity of 26.4 m/s at a 45° angle from the top of a cliff that is 10.0 m above the water below. Use g = 9.80 m/s2 to simplify the calculations. Note: you could answer these questions using projectile motion methods, but try using an energy conservation approach instead.
(a) What is the ball’s speed when it hits the water?
m/s
(b) What is the ball’s speed when it reaches its maximum height?
m/s
(c) What is the maximum height (measured from the water) reached by the ball in its flight?
m

Explanation / Answer

Here vertical component of velocity,v = 26.4sin 45 = 18.66 m/sec

Here Gain in potential energy = loss in kinetic energy

0.5*m*v^2 = mgh

Thus h = v^2/2g = 17.76 m

From water height = 17.76 + 10 = 27.76 m

Here there is no acceleration in the horizontal direction


Thus horizontal component when it hits the ground = 26.4cos45 = 18.66 m/sec


Let the vertical velocity = v m/sec


Apply third equation of motion

v^2 - u^2 = 2*g*s


v^2 = 18.66^2 +2*9.8*10 = 544.195 m/sec

Thus v = 23.32 m/sec

At the maximum height its vertical velocity will become Zero

So it has only horizontal velocity = 18.66 m/sec

Thus resultant velocity = 18.66^2 + 23.32^2 = 29.86 m/sec

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