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phy201
Your 'cq_1_08.2' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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Asst 8 Question 1
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A ball is tossed upward at 15 meters / second from a height of 12 meters above the ground. Assume a uniform downward acceleration of 10 m/s^2 (an approximation within 2% of the 9.8 m/s^2 acceleration of gravity).
• How high does it rise and how long does it take to get to its highest point?
answer/question/discussion: ->->->->->->->->->->->-> :
The given
The ball original height (x0) is 12m, v0 = 15 m/s, a =10 m/s^2
All force acting downward will be considered negative and all forces acting upwards will be considered positive, so the acceleration is -10m/s^2.
Recall that ds is the difference in distance = xf-x0 where xf is the finial position and x0 is the initial position of an object.
When the ball reaches its maximum height the finial velocity will be 0 so vf = 0
The height the ball reaches
From the equation of motion vf^2 = v0^2 + 2 a `ds.
Rearranging this equation to make ds the subject of the equation will give the result.
ds = (vf^2 - v0^2 ) / (2*a)
ds = ((0m/s) ^2 - (15 m/s)^2 ) / (2*-10 m/s^2)
ds = (-225 (m/s )^2) / (-20 m/s^2)
ds = 11.25 m
The ball travels 11.25 m above its original position with is 23.25m (ds + x0 = 11.25+12) above the ground.
The time it takes to reach maximum height of 11.25m.
From the equation of motion vf = v0 + a * `dt
Rearrange the equation to make dt the subject of the equation give us
dt = (vf-v0) / a
dt = (0 m/s - 15 m/s) / (-10 m/s^2)
dt = -15m/s / (-10 m/s^2)
dt = 1.5 s
It takes the ball 1.5 seconds to reach its maximum height
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• How fast is it then going when it hits the ground, and how long after the initial toss does it first strike the ground?
answer/question/discussion: ->->->->->->->->->->->-> :
First keep in mind that all forces and measurements going up are considered positive and all forces and measurements acting downward are considered negative, so a = -10 m/s^2 and ds = -12m ( because the ball was tossed from 12 m above the ground.)
v0 = 15m/s, ds = -12m, a = -10 m/s^2
The velocity when it hits the ground
Recall from the equations of motion vf^2 = v0^2 + 2 a `ds.
vf^2 = v0^2 + 2 a `ds.
vf^2 = (15 m/s)^2 + 2 * (-10m/s^2) * (-12m)
vf^2 = 225 (m/s)^2 + 240 (m/s)^2
vf^2 = 465 (m/s)^2
vf = sqrt(465 (m/s)^2)
vf = 21.56 m/s or vf = -21.56 m/s (since the ball will be travelling downwards the correct velocity is -21.56 m/s with the negative sine indicating that the ball is travelling downward)
when the ball reaches the ground it has a velocity of -21.56 m/s
The time taken for the ball to hit the ground
Recall from the equations of motion `ds = v0 `dt + .5 a `dt^2
ds = v0*dt +1/2 *a*dt^2
-12 m = 15 m/s *dt + ½ * (-10 m/s^2) * dt^2
-12 m = 15 m/s *dt + (-5 m/s^2)*dt^2
(-5 m/s^2)*dt^2 + 15 m/s *dt + 12m = 0 (this is a quadratic equation so we use the formula to solve )
recall if ax^2 + bx +c =0 then x = [-b +(sqrt(b^2 - 4*a*c))]/(2*a) or x = [-b - (sqrt(b^2 - 4*a*c)) ]/(2*a)
in our problem a = -5, b = 15, c = 12 and dt = x
dt =[ -15 + (sqrt(15^2 - 4*(-5)*12))]/ (2*(-5))
dt = [-15 + 21.56] / (-10)
dt = -0.656 s (since we are calculating time this value cannot be correct.)
or
dt =[ -15 - (sqrt(15^2 - 4*(-5)*12))] / (2*(-5))
dt = [-15 - 21.56] / (-10)
dt = 3.66 s
So the ball will reach the ground in 3.66 seconds after it is tossed into the air.
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• At what clock time(s) will the speed of the ball be 5 meters / second?
answer/question/discussion: ->->->->->->->->->->->-> :
Identify the give; v0 = 15 m/s, vf = 5 m/s, a = -10 m/s^2
Recall from the equation of motion vf = v0 + a * `dt. If we rearrange this equation to make dt the subject.
dt = (vf - v0)/a
dt = (5 m/s - 15 m/s) / (-10 m/s^2)
dt = -10 m/s / -10 m/s^2
dt = 1 s
The ball will have a velocity of 5 m/s one second after it is tossed up in the air.
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• At what clock time(s) will the ball be 20 meters above the ground?
• How high will it be at the end of the sixth second?
answer/question/discussion: ->->->->->->->->->->->-> :
clock time(s) will the ball be 20 meters above the ground
If the ball is 20 m above the ground then it is 8m (20m -12m) above the point at which it was tossed.
recall ds = 8 m, v0 = 15 m/s, a = -10 m/s^2
Recall from the equations of motion `ds = v0 `dt + .5 a `dt^2
ds = v0*dt +1/2 *a*dt^2
8 m = 15 m/s *dt + ½ * (-10 m/s^2) * dt^2
8 m = 15 m/s *dt + (-5 m/s^2)*dt^2
(-5 m/s^2)*dt^2 + 15 m/s *dt - 8m = 0 (this is a quadratic equation so we use the formula to solve )
recall if ax^2 + bx +c =0 then x = [-b +(sqrt(b^2 - 4*a*c))]/(2*a) or x = [-b - (sqrt(b^2 - 4*a*c)) ]/(2*a)
in our problem a = -5, b = 15, c = -8 and dt = x
dt =[ -15 + (sqrt(15^2 - 4*(-5)*(-8)))]/ (2*(-5))
dt = [-15 + 8.06] / (-10)
dt = 0.69 s
or
dt =[ -15 - (sqrt(15^2 - 4*(-5)*(-8)))] / (2*(-5))
dt = [-15 - 8.06] / (-10)
dt = 2.31 s
So the ball will reach a height of 20m above the ground at 0.69s and 2.31s after it is tossed into the air.
The high will it be at the end of the sixth second
Identify the give; v0 = 15 m/s, dt = 6 s, a = -10 m/s^2
Recall from the equation of motion `ds = v0 `dt + .5 a `dt^2
ds = v0 * dt + ½ * a * dt^2
ds = 15 m/s * 6 s + ½ * (-10 m/s^2) * (6 s)^2
ds = 90m -180m
ds = - 90m
Since the ball was tossed at a height of 12m above ground the ball will be on the ground after 6 s.
.
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