Question

A man stands on the roof of a building of height 14.4 m and throws a rock with a velocity of magnitude 33.2 m/s at an angle of 27.5 ∘above the horizontal. You can ignore air resistance.

Part-A Calculate the maximum height above the roof reached by the rock.

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

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

Answer 1

(A) v0x = 33.2 cos27.5 = 29.45 m/s

and v0y = 33.2 sin27.5 = 15.33 m/s

at maximum height, vertical component of velocity will become zero.

vy = 0

Applying , vf^2 - vi^2 = 2 a d

0^2 - (15.33)^2 = 2(-9.8)(H)

H = 12m .........Ans

(B) horizontal component will remain same.

vx = v0x = 29.45 m/s

in vertical, (initial position to the ground)

vf^2 - vi^2 = 2(a)(yf - yi)

vy^2 - (15.33)^2 = 2(-9.8)(0 - 14.4)

vy = - 22.7 m/s

magnitude = sqrt(vx^2 + vy^2) = 37.2 m/s ......Ans

(C) in vertical, vf = vi + a t

-22.7 = 15.33 + (-9.8)t

t = 3.88 sec

In horizontal, X = (vx) t

X = 114.3 m