A rock is projected from the edge of the top of a building with an initial veloc

Question

A rock is projected from the edge of the top of
a building with an initial velocity of 18.8 m/s
at an angle of 34 above the horizontal. The
rock strikes the ground a horizontal distance
of 73 m from the base of the building.
The acceleration of gravity is 9.8 m/s2 .
Assume: The ground is level and that the
side of the building is vertical. Neglect air
friction.
1. What is the vertical component of the rock’s
velocity when it strikes the ground?
Answer in units of m/s

2. What is the magnitude of the rock’s velocity
when it strikes the ground?
Answer in units of m/s

Answers are NOT 10.82 or 15.85 for number 1, and NOT 18.8 for number 2.
I tried them

the horizonntal component of the rocks velocity as it hits the ground is 15.58 m/s, the rock is airborn for 4.69 seconds, and the building is 58 meters tall

Explanation / Answer

ucos34*t =73 t =4.68 secs Vy =usin34-g(t) = usin34-9.8*4.68 = -35.35 m/s Vx =ucos34 =15.585 m/s V =vVy2+Vx2 =38.63 m/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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