a Ball is thrown upwards from a roof of a building at 20m/s at 35degrees above the horizontal
A) what maximum hight does it reach above the launch point (i got it to be 6.21m not sure if I'm correct)
B) there is a second building 50m away. How high relative to the launch point is the ball when it hits the building? (i got it to be 44.19m below the launch point, again not sure if this is correct.)
A) what maximum hight does it reach above the launch point (i got it to be 6.21m not sure if I'm correct)
B) there is a second building 50m away. How high relative to the launch point is the ball when it hits the building? (i got it to be 44.19m below the launch point, again not sure if this is correct.)
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Use Vf^2 = Vi^2 + 2*a*d where Vf = 0 at peak altitude when the ball's vertical velocity slows to a stop just before falling at acceleration g. Vi is the initial vertical velocity, a = -9.8m/s^2 and d is the peak altitude.
0 = [20sin35]^2 - 19.6*d => d = 6.7m above the roof of the building
Because we are ignoring air resistance, the horizontal component of velocity Vx is constant
Vx = 20cos35 = 16.4 m/s. How long does it take the ball to travel 50m at 16.4m/s
t = 50/16.4 = 3.1s What is the vertical location of the ball at t = 3.1s?
Let the initial height = Yo
Y(t) = Yo + Vyi*t - 4.9t^2
Y(3.1) = Yo + 20sin35*3.1 - 4.9*3.1^2 = Yo - 10.6 so the ball hits the building 10.6m below the launch height.
0 = [20sin35]^2 - 19.6*d => d = 6.7m above the roof of the building
Because we are ignoring air resistance, the horizontal component of velocity Vx is constant
Vx = 20cos35 = 16.4 m/s. How long does it take the ball to travel 50m at 16.4m/s
t = 50/16.4 = 3.1s What is the vertical location of the ball at t = 3.1s?
Let the initial height = Yo
Y(t) = Yo + Vyi*t - 4.9t^2
Y(3.1) = Yo + 20sin35*3.1 - 4.9*3.1^2 = Yo - 10.6 so the ball hits the building 10.6m below the launch height.