(a) Find the speed at which the ball was launched. (Give your answer to two deci
ID: 1656512 • Letter: #
Question
(a) Find the speed at which the ball was launched. (Give your answer to two decimal places to reduce rounding errors in later parts.)
1 m/s
(b) Find the vertical distance by which the ball clears the wall.
2 m
(c) Find the horizontal distance from the wall to the point on the roof where the ball lands.
3 m
Explanation / Answer
(a) Here, set the equation of motion in the vertical as y=0 at the surface of the playground
Use the expression -
s = u*t + (1/2)*a*t^2
Now, for the ball
y(t) = -4.6 + v0*sin(53)*t - 0.5*9.81*t^2
Again, we are given that
24 = v0*cos(53)*2.20
=> v0 = 24 / [2.20*cos(53)] = 18.13 m/s.
(b) Now, check with y(t) to see if the ball is above the wall
y(2.2) = -4.6 + 18.13*sin(53)*2.2 - 0.5*9.81*2.2^2
=> y(2.2) = -4.6 + 31.85 - 23.74 = 3.51 m
This is 3.51 m above the wall.
(c) Now, let's find t for when y(t)=0
-4.6 + v0*sin(53)*t - 0.5*9.81*t^2 = 0
=> -4.6 + 18.13*0.80*t - 4.9*t^2 = 0
=> 4.9*t^2 - 14.5t + 4.6 = 0
=> t = [14.5 + sqrt{14.5^2 - 4*4.9*4.6}] / (2*4.9) = [14.5 + 10.96] / 9.8 = 2.60 s
The smaller value of t is when the ball reaches the plane of the playground on the ascent. So, you can discard this.
So, t = 2.60 seconds
In that time the ball has traveled
x(2.60) = v0*cos53*2.60 = 18.13* 1.56 = 28.37 m
So, the horizontal distance from the wall to the point on the roof where the ball lands = 28.37 - 24.0 = 4.37 m
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