Answer to Question #144417 in Physics for Tristan

Question

A person standing on top of a 40.0m high building throws a ball with an initial velocity of 25.0m/s at an angle of 20.0° below horizontal. How far from the base of the building with the ball land?

Answer 1

Let's write the equations of motion of the ball in horizontal and vertical directions:

$$x=v_0tcos\theta, (1)$$ $$y=y_0+v_0tsin\theta+\dfrac{1}{2}gt^2, (2)$$

here, $x$ is the horizontal displacement of the ball (or the range of the ball), $v_0=25.0\ \dfrac{m}{s}$ is the initial velocity of the ball, $t$ is the total flight time of the ball, $\theta=20^{\circ}$ is the launch angle, $y$ is the vertical displacement of the ball (or the height), $y_0=40.0\ m$ is the initial height from which the ball was thrown and $g=-9.8\ \dfrac{m}{s^2}$ is the acceleration due to gravity.

Let's first find the total flight time from the second equation (the vertical component of the ball's velocity directed downward, so it will be with sign minus):

$$0=40-25t\cdot sin20^{\circ}-\dfrac{1}{2}\cdot 9.8t^2,$$ $$4.9t^2+8.55t-40=0.$$

This quadratic equation has two roots:

$$t_1=\dfrac{-b+\sqrt{b^2-4ac}}{2a}=\dfrac{-8.55+\sqrt{(-8.55)^2+784}}{2\cdot 4.9}=2.11\ s,$$ $$t_2=\dfrac{-b-\sqrt{b^2-4ac}}{2a}=\dfrac{-8.55-\sqrt{(-8.55)^2+784}}{2\cdot 4.9}=-3.86\ s.$$

Since time can't be negative, the correct answer is $t=2.11\ s.$

Finally, we can substitute the total flight time of the ball into the first equation and find the range of the ball:

$$x=v_0tcos\theta=25.0\ \dfrac{m}{s}\cdot 2.11\ s\cdot cos20^{\circ}=49.6\ m.$$

Answer:

$x=49.6\ m.$