A man stands on the roof of a 21.0-m tall building and throws a rock with a velo

Question

A man stands on the roof of a 21.0-m tall building and throws a rock with a velocity of magnitude 22.0 m/s at an angle of 34.0 degree above the horizontal. You can ignore air resistance. Calculate the maximum height above the roof reached by the rock. m Calculate the magnitude of the velocity of the rock just before it strikes the ground. m/s Calculate the horizontal range from the base of the building to the point where the rock strikes the ground. m Draw x - t, y - t, v_x - t and v_y - t graphs for the motion. (Do this on paper. Your instructor may ask you to turn in these graphs.)

Explanation / Answer

h = 21 m ; Vi = 22 m/s ; theta = 34 deg

a)Vix = 22 x cos34 = 18.24 m/s

Viy = 22 x sin34 = 12.30 m/s

we know that, v = u + at

at max height, v = 0

0 = 12.3 - 9.8 x t => t = 1.26 s

s = h + ut + 1/2 at^2

s = 21 + 12.3 x 1.26 - 1/2 x 9.8 x 1.26^2 = 28.72 m

this is the height above the ground. the height above the building is:

H = 27.72 - 21 = 7.72 m

Hence, H = 7.72 m

b)at ground,

0 = 12.3 x t - 1/2 x 9.8 x t^2 + 21

t = 3.68 s

vy = 12.3 - 9.8 x 3.68 = -23.76 m/s

v = sqrt (23.76^2 + 18.24^2) = 29.95 m/s

Hence, v = 29.95 m/s

c)D = Vx x t

D = 18.24 x 3.68 = 67.12 m

Hence, D = 67.12 m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.