Question

Lacie kicks a football from ground level at a velocity of 13.9 m/s and at an angle of 25.0° to the ground. You have determined that the football would travel 15.1 m before landing. How would this value change if the football was kicked at an angle of 35.0°?Complete all equations without rounding and then round to the nearest tenth at the end.

Answer 1

The horizontal distance traveled by the ball (range of motion) is given by the following equation:

$x= v_{0x} t$

In which $x$ is the range of motion, $v_{0x}$ is the horizontal component of the initial velocity, and $t$ is the time of motion.

First, lets calculate the horizontal component of the initial velocity:

$v_{0x}= v_{0}cos( \alpha)=13.9cos(35)=11.39$

Now, we calculate the time of motion from the equation that described the motion of the ball in the vertical axis:

$y= \frac{1}{2}at^2+ v_{0y}t$

In which $y$ is the position of the ball vertically, $v_{0y}$ is the vertical component of the initial velocity, $a$ is the acceleration in the vertical axis (which is gravity), and $t$ is time of motion.

We want to find the time when the ball lands, hence, when $y=0$; so the equation becomes:

$0= \frac{1}{2}at^2+ v_{0y}t=\frac{1}{2}at^2+ v_{0}sin( \alpha )t=\frac{1}{2}(-9.8)t^2+ 13.9sin(35 )t$

We rewrite it a bit more:

$-4.9t^2+7.97t=0$

This is a quadratic equation, so we use the quadratic equation formula to solve for time (we'll get two answers):

$t= -\frac{1}{9.8} [{-7.97}+-\sqrt{(7.97)^2}]$

Clearly, one of the answers is $t=0$, this is before you kick the ball (it is on the ground), we want the nonzero answer (when it lands) so:

$t= -\frac{1}{9.8} ({-7.97}-7.97})=1.63$

Now, we plug-in the time value to the equation of the motion's range:

$x= v_{0x} t=(11.39)(1.63)=18.57$

The ball will travel 18.57 meters.

Answer 2

If the football is kicked at a 35.0º angle, it will travel 18.5 m before landing. This is farther than if it were kicked at a 25° angle.