A man stands on the roof of a building of height 14.3 mand throws a rock with a

Question

A man stands on the roof of a building of height 14.3 mand throws a rock with a velocity of magnitude 30.0 m/sat an angle of 29.9 above the horizontal. You can ignore air resistance.

Part A

Calculate the maximum height above the roof reached by the rock.

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Part B

Calculate the magnitude of the velocity of the rock just before it strikes the ground.

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Part C

Calculate the horizontal distance from the base of the building to the point where the rock strikes the ground.

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A man stands on the roof of a building of height 14.3 mand throws a rock with a velocity of magnitude 30.0 m/sat an angle of 29.9 above the horizontal. You can ignore air resistance.

Part A

Calculate the maximum height above the roof reached by the rock.

y =   m  

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Part B

Calculate the magnitude of the velocity of the rock just before it strikes the ground.

v =   m/s  

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Part C

Calculate the horizontal distance from the base of the building to the point where the rock strikes the ground.

x =   m  

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Explanation / Answer

as g=9.8m/s^2
(a) h=v^2/2g
h= (30*sin(29.9))^2/(2*9.8) = 11.41 m
h= 14.3 + 11.41 = 25.71 [m]

(b) v=(2gh) = (2*9.8*25.71)= 22.45 m/s (vertical)
v= 30*cos(29.9)= 26 (horizontal)
v= (22.45^2+26^2)= 34.35 [m/s]

(c) t1=v/g = (30*sin(29.9))/9.8= 1.53s (Time to highest point)
t2=(2h/g) = (2*25.71/9.8)= 2.3s (Time to the ground)
t=t1+t2 = 1.53 +2.3= 3.83s
d=vt = 30*cos(29.9)*3.83= 99.6[m] approx

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