+64 A ball is thrown upward with an initial velocity of 50 feet per second. The height h (in feet) of the ball after t seconds is given by h = -16t2 +50t. At the same time, a balloon is rising at a constant rate of 20 feet per second. Its height h in feet after seconds is given by h = 20t.
a. When do the ball and the balloon reach the same height?
b. When does the ball reach it's maximum height?
c. When does the ball hit the ground?
+129840 a. To see when they reach the same height, set the equations equal
-16t^2 + 50t = 20 t subtract 20t from each side
-16t^2 + 30t = 0 factor
-2t [ 8t - 15] = 0 set each factor = 0
-2t = 0 → t = 0 [ trivial answer ]
8t - 15 = 0 → t = 15/8 seconds = 1.875 seconds
b. To find the time when the ball reaches it's maximum height, we have
- 50 / [ 2 (-16) ] = t = -50/-32 = 1.5625 seconds
c. Because of the symmetric nature of the function, the ball hits the ground in twice the time it takes to reach its max height = 2(1.5625) seconds = 3.125 seconds
