A ball is thrown straight upward at an initial speed of v0 = 80 ft/s. (Use the formula h = −16t2 + v0t . If not possible, enter IMPOSSIBLE.) (a) When does the ball initially reach a height of 64 ft? (b) When does it reach a height of 113 ft? (c) What is the greatest height reached by the ball? (d) When does the ball reach the highest point of its path? (e) When does the ball hit the ground?

Question

shamim · Answer

a)  h=v0*t-(1/2)g*t^2
Here h=64ft
v0=80ft
t=?

radar · Answer

I would suggest using the equation that the problem requested you use.  It will get the same results as the formula presented by shamim.

radar · Answer

substituting values that were provided by the problem for (a) you should get something like this
64 = -16t^2 + 80 t     Rearrange it to the familiar quadratic form:
16t^2 - 80t + 64 = 0  Divide thru by 16 getting:

radar · Answer

t^2 - 5t + 4 = 0       Use the smaller value as it asked for the initial 64, it will reach that height twice once on the way up and once on the way down,  Good luck.