Question
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.
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.
Answers
the same as when it went up.
oops. not so, due to h0
so, find t when h(t)=0
then use that to get v(t) at that time.
so, find t when h(t)=0
then use that to get v(t) at that time.
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