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UkoKoshka [18]
3 years ago
10

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
Mathematics
1 answer:
satela [25.4K]3 years ago
4 0

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

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