when h is zero it hit the ground.so put h=0 and solve for t.The height, h, in feet of a golf ball above the ground after being hit into the air is given by the equation, h = -16t^2 + 64t, where t is the number of seconds elapsed since the ball was hit. How many seconds does it take for the golf ball to hit the ground? The height, h, in feet of a golf ball above the ground after being hit into the air is given by the equation, h = -16t^2 + 64t, where t is the number of seconds elapsed since the ball was hit. How many seconds does it take for the golf ball to hit the ground?
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h = -16t^2 + 64tIt hits the ground when h=0.
-16t^2 + 64t = 0
Solve for tThe ball is on the ground whenever the height is 0:
Obviously, is the time when the ball is thrown, so ignore that. It thus takes the ball seconds to hit the ground after it's been launched. -16t^2+64=0 ----------> 16t(t-4)=0---------> t=0, t=4t=0
is the time when the ball is thrown, so ignore that. It thus takes the ball thus t=4 seconds to hit the ground after the ball is launched. So with that being said it took 4 seconds for the ball to hit the ground or 0 seconds. 0-4 Hope this helps!
