A baseball thrown at an angle of 55.0 above the horizontal strikes a building 18

Question

A baseball thrown at an angle of 55.0 above the horizontal strikes a building 18.0 m away at a point 6.00 m above the point from which it is thrown. Ignore air resistance.

Part A

Find the magnitude of the initial velocity of the baseball (the velocity with which the baseball is thrown) (m/s)

Part B

Find the magnitude of the velocity of the baseball just before it strikes the building (m/s)

Part C

Find the direction of the velocity of the baseball just before it strikes the building.(degrees below the horizontal.)

Explanation / Answer

along horizantal

initial velocity vox = vo*cos55

acceleration = ax = 0

displacement x = vox*T

T = x/vox

along vertical

initial velocity voy = vo*sin55

acceleration ay = -g

y = voy*T + 0.5*ay*T^2

T = x/vox

y = tan55*x - 4.9*x^2/(vo^2*(cos55)^2)

for x = 18 m    and y = 6 m

6 = tan55*18 - 4.9*18^2/(vo^2*(cos55)^2)

vo = 15.6 m/s <<-------------answer

++++++++++++

part(B)

T = x/vx = 18/(15.6*cos55) = 2.01 s

vy = voy + ay*T = (15.6*sin55)-(9.8*2.01) = -6.92 m/s

v = sqrt(vox^2+vy^2)

v = sqrt(8.95^2+6.92^2) = 11.3 m/s

drection = tan^-1(vy/vx) = 37.7 degrees below the horizantal

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