The function h(t) = -4.9t^2 + 49t expresses the approximate height, h, in meters, of a ball t seconds after it is launched vertically upward from the ground with an initial velocity of 49 meters per second. After approximately how many seconds will the ball hit the ground?
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It will hit the ground when h = 0
-4.9t^2 + 49t = 0
-4.9t(t - 10) = 0
t = 0 , at the start of the throw,
t = 10 , when it returns to the ground
V = Vo + g*Tr = 0.
49 + (-9.8)Tr = 0,
Tr = 5 s. = Rise time.
Tf = Tr = 5 s. = Fall time.
Tr + Tf = 5 + 5 = 10 s. = Time to reach gnd.
To find the time it takes for the ball to hit the ground, we need to determine when the height, h(t), equals zero.
Given the function h(t) = -4.9t^2 + 49t, we can set it equal to zero:
-4.9t^2 + 49t = 0
Next, we can factor out a common term:
t(-4.9t + 49) = 0
Now, we have two factors: t = 0 and -4.9t + 49 = 0.
First, let's consider the factor t = 0. Since time cannot be negative, this factor does not give us a meaningful solution.
Now, let's solve for the other factor:
-4.9t + 49 = 0
Subtract 49 from both sides:
-4.9t = -49
Divide both sides by -4.9 to isolate t:
t = -49 / -4.9
Simplifying, we have:
t = 10
Therefore, the ball hits the ground after approximately 10 seconds.