A stone is thrown from the top of a building at an angle of 30.0 degree above th

Question

A stone is thrown from the top of a building at an angle of 30.0 degree above the horizontal with an initial speed of 20 m/s. The point of release is 45.0 m above the ground. (a) How long does it take for the stone to hit the ground? (b) What is the magnitude and direction of the stone's acceleration right after it leaves the building? Explain. (c) What is the magnitude and direction of the stone's acceleration right before it touches the ground? (d) Find the magnitude and direction of the stone's velocity right before it touches the ground, (e) Find the horizontal distance that the stone traveled.

Explanation / Answer

Here ,

u = 20 m/s

theta = 30 degree

h = 45 m

a) let the time taken is t

Using second equation of motion

-h = u * sin(theta) * t - 0.50 * g * t^2

- 45 = 20 * sin(30 degree) * t - 0.50 * 9.8 * t^2

solving for t

t = 4.22 s

the time taken is 4.22 s

b)

as the acceleration at every point in freefall will be same

magnitude of acceleration = g = 9.8 m/s^2

direction of acceleration is downwards

c)

as the acceleration at every point in freefall will be same

magnitude of acceleration = g = 9.8 m/s^2

direction of acceleration is downwards

d)

let the final vertical velocity is v

vy = 20 * sin(30 degree) - 9.8 * 4.22 = -31.4 m/s

vx = 20 * cos(30 degree) = 17.32 m/s

magnitude of stone's velocity = sqrt(31.4^2 + 17.32^2)

magnitude of stone's velocity = 35.9 m/s

direction of stone's velocity = arctan(31.4/17.32) below horizontal

direction of stone's velocity = 61.1 degree below horizontal

e)

horizontal distance = horizontal speed * time

horizontal distance = 17.32 * 4.22

horizontal distance = 73.1 m

horizontal distance is 73.1 m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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