5 A man throws a rock with a velocity of 35 m/s at an angle of 25 degree above t

Question

5 A man throws a rock with a velocity of 35 m/s at an angle of 25 degree above the horizontal from the roof of a 60 m tall building. Calculate (a) the horizontal and vertical components of the rock's initial velocity (4 points). (b) the maximum height abuse the ground reached by the rock (8points). (c) the horizontal distance from the base of the building to the point where the rock strikes the ground (12points). Extra points (d) the velocity vector (magnitude and angle above the horizontal) of the rock just before it strikes the ground (3Points).

Explanation / Answer

vox = vo*cos25 = 35*cos25 = 31.72 m/s

voy = vo*sin25 =35*sin25 = 14.79 m/s

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b)H = h + voy^2/2g

H = 60 + (14.79*14.79)/(19.6) = 71.16 m

c)

vertical direction

h = voy*T - 0.5*g*T^2

-60 = 14.79*T - 4.9*T^2

T = 5.32 s

x = vox*T = 31.72*5.32 = 168.7504 m <--------answer

d) vx = vox = 31.72

vy = voy - gT = 14.79-(9.8*5.32) = -37.346 m/s

V = sqrt(Vx^2+vy^2) = sqrt(31.72^2+37.346^2)

V = 48.99 m/s <-----answer

direction = tan^-1(Vy/Vx) = 49.65 below the horizantal

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