In the figure, a ball is thrown leftward from the left edge of the roof, at heig

Question

In the figure, a ball is thrown leftward from the left edge of the roof, at height h above the ground. The ball hits the ground 2.30 s later, at distance d = 20.0 m from the building and at angle = 59.0° with the horizontal. (a) Find h. (Hint: One way is to reverse the motion, as if on videotape.) What are the (b) magnitude and (c) angle relative to the horizontal of the velocity at which the ball is thrown (positive angle for above horizontal, negative for below)?

This answer has no units° (degrees)mkgsm/sm/s^2NJWN/mkg·m/s or N·sN/m^2 or Pakg/m^3gm/s^3times

Explanation / Answer

x = 20 = ucos(tehta)*2.3

y = -h = (usin(theta)*2.3-1/2*9.8*2.3^2

vx = ucos(theta)

vy = usin(theta)-9.8*2.3

tan(59) = vy/vx

ucos(theta) = 20/2.3 = 8.69 m/s

vy = -8.69*tan59 = -14.46 m/s

usin(theta)-9.8*2.3 = -14.46

usin(theta) = 8.08 m/s

now a) h = usin(tehta)*2.3+4.9*2.3^2

h = 8.08*2.3+4.9*2.3^2 = 44.5 m

b) u = sqrt(8.69^2+8.08^2) = 11.87 m/s

c) theta = tan^-1(8.08/8.69) = 42.9 degerees

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