Hi I have another question from my practice physics test I\'d like some help on.

Question

Hi I have another question from my practice physics test I'd like some help on. If you chould walk me through it/show your work that'd be awesome. Thanks!

A stone is thrown at an angle of 60 degree above the horizontal from the top of a building with an initial speed of 12.0 m/s. The stone flies for 5.0 seconds before hitting ground. Ignore air resistance. How far out from the building does the stone travel horizontally when it hits the ground? What is the height of the building? (4 points) Measured from the top of the building, what is the maximum height the stone reached? Calculate the stone's speed right before the impact and the impact angle.

Explanation / Answer

a) horizontally

d = ucos60 x t = 12 x cos60 x 5 = 30m

b) in vertical , using h= ut + at^2 /2

h = 12sin60 x 5 - 9.8 x 5^2 /2   =- 70.54 m

c) at top vy = 0

sp using . v^2 - u^2 = 2ah

0 - (12sin60)^2 =2 x -9.8 x h

h   =5.51 m

d)vx = 12cos60 =6 m/s

vy =12sin60 - 9.8x5 = -38.61 m/s

speed = sqrt(vx^2 + vy^2) =39.07 m/s

direction = tan-1(vy/vx) =81.17 degrees below

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