As preparation for this problem, review Conceptual Example 10. The drawing shows

Question

As preparation for this problem, review Conceptual Example 10. The drawing shows two planes each dropping an empty fuel tank. At the moment of release each plane has the same speed of 127 m/s, and each tank is at the same height of 4.77 km above the ground. Although the speeds are the same, the velocities are different at the instant of release, because one plane is flying at an angle of 15.0 above the horizontal and the other is flying at an angle of 15.0 below the horizontal. Find the (a) magnitude and (b) direction of the velocity with which the fuel tank hits the ground if it is from plane A. Find the (c) magnitude and (d) direction of the velocity with which the fuel tank hits the ground if it is from plane B. In each part, give the direction as a positive angle with respect to the horizontal.

As preparation for this problem, review Conceptual Example 10. The drawing shows two planes each dropping an empty fuel tank. At the moment of release each plane has the same speed of 127 m/s, and each tank is at the same height of 4.77 km above the ground. Although the speeds are the same, the velocities are different at the instant of release, because one plane is flying at an angle of 15.0½ above the horizontal and the other is flying at an angle of 15.0½ below the horizontal. Find the (a) magnitude and (b) direction of the velocity with which the fuel tank hits the ground if it is from plane A. Find the (c) magnitude and (d) direction of the velocity with which the fuel tank hits the ground if it is from plane B. In each part, give the direction as a positive angle with respect to the horizontal.

Explanation / Answer

for plane flying 15 degrees above horizontal
Vx =127cos15 =122.67 m/s
Vy =127sin15 =32.87

v^2 =u^2 +2as =32.87^2 +2*4770*9.8
hence y component of velocity v = - 307.53 m/s
hence magnitude of final velocity
=(122.67^2+307.53^2) = 331.09 m/s

tan = - 122.67/307.53

hence = 338.25 degrees from +ve x direction

.............

exactly same answer for plane flying 15 degrees below horizontal

as u^2 term cannot differntiate between +ve or -ve u

hence for plane flying 15 degrees below horizontal

for plane flying 15 degrees above horizontal
Vx =127cos15 =122.67 m/s
Vy =-127sin15 =-32.87

v^2 =u^2 +2as =32.87^2 +2*4770*9.8
hence y component of velocity v = - 307.53 m/s
hence magnitude of final velocity
=(122.67^2+307.53^2) = 331.09 m/s

tan = - 122.67/307.53

hence = 338.25 degrees from +ve x direction

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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