A ball is thrown horizontally from the top of a building 49.5 m high. The ball strikes the ground at a point 43.5 m from the base of...

Saturday, February 27, 2016

A ball is thrown horizontally from the top of a building 49.5 m high. The ball strikes the ground at a point 43.5 m from the base of...

First, we will calculate the time it takes for the ball to
hit the ground:

To do this, we will use the equation $\displaystyle{y}=$
y_0 + v_0t + 1/2at^2 $\displaystyle.$

In this case, the initial velocity
in the y direction is zero, as is the initial y position, and acceleration here is
$\displaystyle{9.81}$ $\displaystyle\frac{{m}}{{{s}}^{{2}}}$ :

$\displaystyle{49.5}{s}=\frac{{1}}{{2}}\cdot{9.81}\frac{{m}}{{{s}}^{{2}}}\cdot$
t^2

Therefore $t = sqrt{(2*49.5m) / (9.81m/s^2)} =$
3.176s

Now that we know t, we once again use the same
equation, but this time $\displaystyle{a}={0},{x}_{{0}}={0}$ and $\displaystyle{v}_{{0}}$ is the unknown: $\displaystyle{x}=$
v_0*t

Therefore $\displaystyle{v}=\frac{{x}}{{t}}=\frac{{{43.5}{m}}}{{{3.176}{s}}}={13.96}$
m/s

Help Notes

Saturday, February 27, 2016