Question

A ball is thrown horizontally from the top of a 64.5-m building and lands 103.8 m from the base of the building. Ignore air resistance. (Assume the ball is thrown in the +x direction and upward to be in the +y direction.) How long (in s) is the ball in the air?

Answer 1

Answer:

The ball is in the air for 3.628 sec

Explanation:

We can find speed in the "y" direction when the ball lands with this equation:

$V_{fy} ^{2} = V_{oy} ^{2} + 2*g*(y_{f} - y_{o})$

Where:   $V_{oy} = 0 m/sec$   because the ball is thrown horizontally

              $g = - 9.8 m/sec^{2}$   because gravity is a vector that points down in the "y" direction

              $y_{o} = 64.5 m$

              $y_{f} = 0 m$   because the ball lands on the ground

Then:      $V_{fy} ^{2} = (0 m/sec)^{2} + 2(-9.8 m/sec^{2})(0 m - 64.5 m)$

              $V_{fy} ^{2} = 2(- 9.8 m/sec^{2})(- 64.5 m)$

              $V_{fy} ^{2} = 1264.2 m^{2}/sec^{2}$

              $V_{fy} = \sqrt{1264.2 m^{2}/sec^{2}}$

              $V_{fy} = - 35.5556 m/sec$   because the ball is falling and that means speed in the "y" direction is a vector that points down

Now we can calculate the time with next equation:

$V_{fy} = V_{oy} + g*t$

$V_{fy} - V_{oy} = g*t$

$t = \frac{V_{fy} - V_{oy}}{g}$

Then:       $t = \frac{-35.5556 m/sec - 0 m/sec}{- 9.8 m/sec^{2}}$

Finally      t = 3.628 sec