Question

A ball is launched from the top of a building 80 m high at an angle of above the horizontal with a speed of 63 in/s. How long does it take before the ball reaches its highest point? What is the maximum height of the ball above the ground? How long is the ball in the air before it hits the ground? How far does the ball travel horizontally before it hits the ground? What is the acceleration of the ball just before it hits the ground? Extra Credit: What is the velocity of the ball just before it hits the ground?

Answer 1

a.-

To reach the highest point, the final speed should be zero, so:

$v_{f} = v_{0} - gt$

$0 = v_{0} - gt$

$v_{0} = gt$

$t = \frac{v_{0}}{g}$

$t = \frac{(63m/s)sin50}{9.8m/s^{2}}$

ANSWER $t = 4.92s$

b.-

$v{_{f}}^{2} = v{_{0}}^{2} - 2gy$

$v{_{0}}^{2} = 2gy$

$y = \frac{v{_{0}}^{2}}{2g}$

$y = \frac{\left [ (63m/s)sin50 \right ]^{2}}{2(9.8m/s^{2})}$

$y = 118.83m$

And the maximum height above the ground is:

$y = 118.83m + 80m$

ANSWER $y = 198.83m$

c.-

The time to complete the parable movement is:

$t = 2t_{max}$

$t = 2(4.92s)$

$t = 9.84s$

And the speed at this time is:

$v_{f} = v_{0} + gt$

$v_{f} = 0 + (9.8m/s)(4.92s)$

$v_{f} = 48.22m/s$

And the time from here to hit the ground is:

$v_{f} = v_{0} + gt$

$t = \frac{v_{0}}{g}$

$t = \frac{48.22m/a}{9.8m/s^{2}}$

$t = 4.92s$

The complete time is:

$t = 4.92s + 9.84s$

ANSWER $t = 14.76s$

d.-

The horizontal component of the movment is constant, so:

$x = v_{x}t$

$x = (63m/s)cos50(9.84s)$

ANSWER $x = 398.48m$

e.-

ANSWER: When the ball reaches its maximum height the speed is zero. So, from here, it starts a free fall movement until it hits the ground. So the acceleration at this time is the gravity.

I hope it helps!!