In physics, one learns that the height of an object thrown upward from an initial height of h0 feet with an initial velocity of v0 (in feet per second) is given by the formula
h(t)=−16t^2+v0t+h0 feet
where t is the amount of time in seconds after the ball was thrown. Also, the velocity of the object is given by
v(t)=−32t+v0. feet per second
When one uses the metric system, the equations become
h(t)=−4.9t^2+v0t+h0 meters
and
v(t)=−9.8t+v0 meters per second.
An object is projected upward from a height of 33 feet at a velocity of 97 feet per second.
Find the velocity of the object when it hits the ground.
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To find the velocity of the object when it hits the ground, we need to determine the time it takes for the object to reach the ground first.
Given:
Initial height, h0 = 33 feet
Initial velocity, v0 = 97 feet per second
Acceleration due to gravity, g = -32 feet per second squared (negative sign indicates downward direction)
To find the time it takes for the object to reach the ground, we set h(t) equal to zero, since the object hits the ground when the height is zero.
0 = -16t^2 + v0t + h0
Substituting the given values:
0 = -16t^2 + 97t + 33
Now, we need to solve this quadratic equation for t. We can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
In this equation, a = -16, b = 97, and c = 33.
Plugging in these values, we get:
t = (-(97) ± √((97)^2 - 4(-16)(33))) / (2(-16))
Simplifying further, we have:
t = (-97 ± √(9409 + 2112)) / -32
t = (-97 ± √(11521)) / -32
t = (-97 ± 107.33) / -32
Now, we have two possible values for t: t1 and t2. One value will be negative, which we discard since time cannot be negative in this context.
t1 = (-97 + 107.33) / -32
t1 = 10.33 / -32
t1 ≈ -0.323 seconds (discard as negative)
t2 = (-97 - 107.33) / -32
t2 = -204.33 / -32
t2 ≈ 6.385 seconds
Therefore, the object takes approximately 6.385 seconds to hit the ground.
Now that we have the time at which the object hits the ground, we can use the formula for velocity, v(t), to find the velocity of the object when it hits the ground.
v(t) = -32t + v0
Substituting the values:
v(t) = -32(6.385) + 97
v(t) ≈ -204.32 + 97
v(t) ≈ -107.32 feet per second
Therefore, the velocity of the object when it hits the ground is approximately -107.32 feet per second.