SOLUTION: 4. How far does a ball, hit from a height of 3 feet at a speed of 120 feet per second and an angle of 30 degrees, go before it hits the ground? Round your answer to the nearest int

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Question 971350: 4. How far does a ball, hit from a height of 3 feet at a speed of 120 feet per second and an angle of 30 degrees, go before it hits the ground? Round your answer to the nearest integer?
X=(V cos() )t
Y= h+(V sin () )t-16t^2
V IS INITIAL VELOCITY. ()= number for degree
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
h=3= initial height of the ball above the ground (in feet)
V=120= initial ball speed (in feet per second)
g=32ft%22+%2F%22second%5E2= acceleration of gravity
theta=30%5Eo= initial angle of the ball trajectory with the horizontal ground
t= time the ball is in the air (in seconds)
X=%28V%2Acos%28theta%29%29%2At= horizontal displacement of the ball (in feet)
Y=h%2B%28V%2Asin%28theta%29%29%2At-16t%5E2= height of the ball above the horizontal ground (in feet)

When the ball hits the horizontal ground, Y=0 . Then,
h%2B%28V%2Asin%28theta%29%29%2At-16t%5E2=0 , and substituting the known values,
3%2B%28120%2Asin%2830%5Eo%29%29%2At-16t%5E2=0--->3%2B%28120%2A%281%2F2%29%29%2At-16t%5E2=0--->3%2B60t-16t%5E2=0--->16t%5E2-60t-3=0
The solution to that quadratic equation that makes sense for this situation is
t=aproximately3.8 ,
because the other solution is negative
Plugging that value for t ,
and the known values for theta and V into
X=%28V%2Acos%28theta%29%29%2At we get
X=%28120%2Acos%2830%5Eo%29%29%2A3.8--->X=approximately395