Question

a ball is thrown with a slingshot at a velocity of 110ft/sec at an angle of 20 degrees above the ground from a height of 4.5 ft. approximentaly how long does is take for the ball to hit the ground. Acceleration due to gravity is 32ft/s^2

Answer 1

Answer:

$t=2.47\ s$  

Step-by-step explanation:

The equation that models the height of the ball in feet as a function of time is

$h(t) = h_0 + s_0t -16t ^ 2$

Where $h_0$ is the initial height, $s_0$ is the initial velocity and t is the time in seconds.

We know that the initial height is:

$h_0 = 4.5\ ft$

The initial speed is:

$s_0 = 110sin(20\°)\\\\s_0 = 37.62\ ft/s$

So the equation is:

$h (t) = 4.5 + 37.62t -16t ^ 2$

The ball hits the ground when when $h(t) = 0$

So

$4.5 + 37.62t -16t ^ 2 = 0$

We use the quadratic formula to solve the equation for t

For a quadratic equation of the form

$at^2 +bt + c$

The quadratic formula is:

$t=\frac{-b\±\sqrt{b^2 -4ac}}{2a}$

In this case

$a= -16\\\\b=37.62\\\\c=4.5$

Therefore

$t=\frac{-37.62\±\sqrt{(37.62)^2 -4(-16)(4.5)}}{2(-16)}$

$t_1=-0.114\ s\\\\t_2=2.47\ s$  

We take the positive solution.

Finally the ball takes 2.47 seconds to touch the ground