A man throws a rock from the top of a cliff that is 55.0 m above ground level wi

Question

A man throws a rock from the top of a cliff that is 55.0 m above ground level with an initial speed of 20.0 m/s at an angle of 38.0° above the horizontal.
(a) Determine the time it takes for the rock to hit the ground below the cliff.
s
(b) Determine the horiztonal distance that the rock travels.
km
(c) At the instant just before the rock hits the ground below the cliff, find the horizontal and the vertical components of its velocity. (Take up and to the right as positive directions.)
horizontal m/s
vertical m/s
(d) What is the the magnitude of the velocity of the rock just before it hits the ground?
m/s
(e) What is the angle made by the velocity vector of the rock just before it hits the ground?
° (below the horizontal)
(f) Find the maximum height above the cliff top reached by the rock.
m

Explanation / Answer

vo=20 m/s==>vox=vo*cos(38º)=15.76 m/s==>voy=vo*sin(38º)=12.31 m/s

cliff ht=55 m==>max ht t=voy/g=1.26 s*2 for rock to be level with cliff at start=2.52 s

-55=voy*t-1/2*g*t²=4.8346 s total time

a) 4.8346 s

b) range=vox*t=76.19 m

c) vx=vox=15.76 m/s==>vyfinal=voy-9.8*4.8346=-35.1 m/s

d) (vox²+vyfinal²)38.445 m/s

e) tan-1(-35.1/15.76)-65.8º

f) max ht above cliff=voy²/(2*g)7.74 m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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