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bumblebeeliz
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two objects thrown in the air. HELP!
A ball is thrown vertically upward with an initial speed of 11 m/s. One second
later, a stone is thrown vertically upward with an initial speed of 25 m/s. (a) Find
the time it takes the stone to catch up with the ball. (b) Find the velocities of the
stone and the ball when they are at the same height.
vo= 11 m/s
vo=25 m/s
v = ?
a/g = -9.80 m/s2
y = ?
yo = 0
t = ?
y= yo + vo*t + 1/2 at^2
Ball Vo= 11m/s
Stone Vo= 25m/s
I have chosen the y=yo+vo*t+1/2 at^2 kinematic equation:
Ball: y= 0 + 11m/s * t + 1/2 (-9.80 m/s2)*t^2
Stone: y= 0 + 25m/s * t + 1/2 (-9.80 m/s2) *t^2
Then I use the Ball=Stone to find the time between the two which equals to:
11m/s * t = 25m/s * t
But this does not get me anywhere. I thought I would eventually use the quadratic formula, but nothing led me to it. I always seem to have trouble when there are two things moving at different times and speeds. Any tips or tricks?
Please help explain :)
Homework Statement
A ball is thrown vertically upward with an initial speed of 11 m/s. One second
later, a stone is thrown vertically upward with an initial speed of 25 m/s. (a) Find
the time it takes the stone to catch up with the ball. (b) Find the velocities of the
stone and the ball when they are at the same height.
vo= 11 m/s
vo=25 m/s
v = ?
a/g = -9.80 m/s2
y = ?
yo = 0
t = ?
Homework Equations
y= yo + vo*t + 1/2 at^2
The Attempt at a Solution
Ball Vo= 11m/s
Stone Vo= 25m/s
I have chosen the y=yo+vo*t+1/2 at^2 kinematic equation:
Ball: y= 0 + 11m/s * t + 1/2 (-9.80 m/s2)*t^2
Stone: y= 0 + 25m/s * t + 1/2 (-9.80 m/s2) *t^2
Then I use the Ball=Stone to find the time between the two which equals to:
11m/s * t = 25m/s * t
But this does not get me anywhere. I thought I would eventually use the quadratic formula, but nothing led me to it. I always seem to have trouble when there are two things moving at different times and speeds. Any tips or tricks?
Please help explain :)
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