Chapter 04, Problem 028 In the figure, a stone is projected at a cliff of height

Question

Chapter 04, Problem 028 In the figure, a stone is projected at a cliff of height h with an initial speed of 41.0 m/s directed at an angle eo 58.0. above the horizontal. The height H reached above after (a) height h of the cif, (b) the speed of the stone just before impact at A, and (c) the maximum the ground. (a) Number (b) Number (c) Number Units Sallow H1N1 LINK TO TEXT LINK TO SAMPLE PROBLE LINK TO SAMPLE PROBLEM LINK TO SAMPLE PROBLEM ivacy Policy I 2000-2017 uohn wiley s Sons, Inc. All Rights Reserved. A Division of lohn wiley sons, unc. Version 4.22.3.4 0 e es a 9 4/23/2017

Explanation / Answer

here,

velocity, v = 41 m/s
time, t = 5.98 s

part a:

From second eqn of motion we have :

height of cliff, h = ho + vy * t - 0.5 * g * t^2
height of cliff, h = 0 + 41*Sin58 * 5.98 - 0.5 * 9.81 * 5.9^2
height of cliff, h = 37.181 m

Part b:
The horizontal motion is steady, so
vx = v0x = v*cos58,

but the vertical component of velocity varies according the equations before.

Thus, the speed at iimpact iis, vh
vh = sqrt((v*cos58)^2 + (v*Sin58 - gt)^2 )
vh = sqrt((41*cos58)^2 + (41*Sin58 - 9.81*5.58)^2 )
vh = 29.51 m/s

Part c:
Max height, H = (v*Sin58)^2/(2g)
Max height, H = (41*Sin58)^2/(2*9.81)
Max height, H = 51.62 m

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

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