Professor's stone

Classical Mechanics Level 3

A 6-ft physics professor, standing on the edge of the a 40-ft high building, throws a stone with a velocity of 50 ft/s at an angle of 37 degrees above the horizontal.

Neglecting air resistance what is the speed of the stone just before it hits the ground?

80.56 73.78 66.35 90.81

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James Guevara by James Guevara , 22, Philippines

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1 solution

James Guevara
Nov 20, 2016

V x = V o c o s θ = 50 c o s 37 Vx = Vocosθ = 50cos37

V y 2 = ( V o s i n θ ) 2 − 2 g y Vy^2=(Vosinθ)^2 - 2gy

V y 2 = ( 50 s i n 37 ) 2 − 2 ( 32 ) ( − 46 ) Vy^2 = (50sin37)^2 -2(32)(-46)

V = √ ( V x ) 2 + ( V y ) 2 V=√(Vx)^2+(Vy)^2

= 73.78 f t / s = 73.78 ft/s

2 Helpful 0 Interesting 0 Brilliant 0 Confused

The angle is actually completely irrelevant to this problem if you just solve it using energy concepts.

Steven Chase - 4 years, 6 months ago

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Without an angle we can solve for the velocity before it hits the ground by using the formula (1/2)(mva^2)+mgh=(1/2)mvd^2 where va is 50 ft/s and vd is the unknown. we can cancel the m and put the values. (1/2)(50^2)+(32)(46)=(1/2)vd^2;vd= √2(1250+1472) ;vd=73.78 ft/s

James Guevara - 4 years, 6 months ago

I also did the same thing. I mistakened with value of g

Md Zuhair - 3 years, 11 months ago

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