# A ball is tossed straight into the air with an initial velocity of 50ft/sec. The ball was 6ft above the ground when thrown. Its height in t is given by, y=-16t^2+50t+6. How high does it go before...

A ball is tossed straight into the air with an initial velocity of 50ft/sec. The ball was 6ft above the ground when thrown. Its height in t is given by, y=-16t^2+50t+6. How high does it go before returning to the ground.

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### 1 Answer

The height of the ball at time time t is given by y=-16t^2+50t+6. We are asked to find the highest point it will reach:

The graph of the function is a parabola that opens down. The maximum height is at the vertex. Given a function of the form f(x)=ax^2+bx+c, the vertex is located at (x,f(x)) where x=-b/(2a).

Thus the t-coordinate of the vertex is t=(-50)/(2*(-16))=50/32=25/16=1.5625

f(t)=45.063

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The maximum height is 45.063 feet, which occurs 1.5625 seconds after the ball is thrown.

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The graph:

** If you have calculus, the maximum occurs where the first derivative is zero.

f(x)=-16t^2+50t+16

f'(x)=-32t+50

f'(x)=0 ==> t=50/32 as before.