Jake serves a volleyball with an initial velocity of 32 feet per second from 4.5 feet above the ground at an angle 0f 35 degrees.
a) Write parametric equations to model the situation.
h(t) = -16t^2 + 32t + 4.5
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b) How long will the ball travel before it hits the ground?
Solve -16t^2+32t+4.5 = 0
t = 2.13 seconds
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parametric equations would be
x = (32 cos 35°) t
y = 4.5 + (32 sin 35°)t - 16t^2
solve for t when y=0
4.5+32t-16t^2 is valid only if the ball is hit straight up.
y(t) = -16 t^2 + 32 sin 35 t + 4.5
x = 32 cos 35
To answer the question, we need to solve the equation -16t^2 + 32t + 4.5 = 0. This equation represents the height of the ball at any given time t.
To solve the equation, we can either use the quadratic formula or factor the equation. Let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In our equation, a = -16, b = 32, and c = 4.5. Plugging these values into the quadratic formula:
t = (-32 ± √((32)^2 - 4(-16)(4.5))) / (2(-16))
Simplifying further:
t = (-32 ± √(1024 + 288)) / -32
t = (-32 ± √(1312)) / -32
t = (-32 ± √(32 * 41)) / -32
t = (-32 ± 8√(41)) / -32
Now, we can simplify the expression:
t = (-32/(-32)) ± (8√(41)/(-32))
t = 1 ± (8√(41)/32)
Now, we have two values for t, which are:
t ≈ 1 - (8√(41)/32)
t ≈ 1 + (8√(41)/32)
To determine the time it takes for the ball to hit the ground, we can disregard the negative value:
t ≈ 1 + (8√(41)/32)
Evaluating this expression:
t ≈ 1.13 seconds
Therefore, the ball will travel for approximately 1.13 seconds before it hits the ground.