Chapter 04, Problem 033 Your answer is partially correct. Try again. A plane, di

Question

Chapter 04, Problem 033 Your answer is partially correct. Try again. A plane, diving with constant speed at an angle of 44.9° with the vertical, releases a projectile at an altitude of 625 m. The projectile hits the ground 6.85 s after release. (a) What is the speed of the plane? (b) How far does the projectile travel horizontally during its flight? What were the magnitudes of the (c) horizontal and (d) vertical components of its velocity just before striking the ground? (State your answers to (c) and (d) as positive numbers.) 176 (d) Number Question Attemptst 2 of S used SAVE FOR LATERSHT

Explanation / Answer

let initial vertical speed of the projectile is v.

acceleration due to gravity=9.8 m/s^2

distance travelled vertically to hit ground=625 m

time taken to hit the ground=6.85 seconds

using the formula:

distance=initial speed*time+0.5*acceleration*time^2

==>625=v*6.85+0.5*9.8*6.85^2

==>v=57.676 m/s

as angle with vertical is 44.9 degrees, then speed of the plane at the time of dropping the projectile=57.676/cos(44.9)

=81.424 m/s

part b:

horizontal component of velocity=81.424*sin(44.9)=57.475 m/s

then horizontal distance travelled=57.475*6.85=393.7 m

part c:

horizontal component of velocity will remain constant at 393.7 m/s.

vertical component of velocity =initial vertical speed+acceleration*time

=57.676+9.8*6.85

=124.81 m/s

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

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