Solve the problem. Unless stated otherwise, assume that the projectile flight is ideal, that the launch angle is measured from the horizontal, and that the projectile is launched from the origin over a horizontal surface
-A projectile is fired at a speed of $720 \mathrm {~m} / \mathrm { sec }$ at an angle of $45 ^ { \circ }$. How long will it take to get $18 \mathrm {~km}$ downrange? Round your answer to the nearest whole number.
A) 33 sec
B) 35 sec
C) 37 sec
D) It will never get that far downrange.

Question

Quizplus · Accepted Answer

The horizontal range $ R $ of a projectile is given by $ R = \frac{v^2 \sin(2\theta)}{g} $. For $ v = 720 \, \text{m/s} $, $ \theta = 45^\circ $, and $ g = 9.8 \, \text{m/s}^2 $, the range is much greater than 18 km. The time $ t $ to reach 18 km is $ t = \frac{18000}{720 \cos(45^\circ)} \approx 35 $ seconds.

Solve the Problem 720 m/sec720 \mathrm {~m} / \mathrm { sec }720 m/sec

Solve the Problem $720 \mathrm {~m} / \mathrm { sec }$