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
-An athlete puts a 16-lb shot at an angle of $39 ^ { \circ }$ to the horizontal from $6.1 \mathrm { ft }$ above the ground at an initial speed of $47 \mathrm { ft } / \mathrm { sec }$. How far forward does the shot travel before it hits the ground? Round your answer to the nearest tenth.
A) 2 ft
B) 74.4 ft
C) 227.8 ft
D) 6.8 ft

Question

Quizplus · Accepted Answer

The distance a projectile travels can be calculated using the formula $d = \frac{v^2 \sin 2\theta}{g}$, where $v$ is the initial velocity, $\theta$ is the launch angle, and $g$ is the acceleration due to gravity. However, because the shot is launched from above the ground, we need to adjust our approach to account for the initial height. The correct formula to use in this scenario involves breaking down the motion into horizontal and vertical components and using the time of flight to find the horizontal distance. The time of flight, given the initial height, can be found using the formula derived from the vertical motion under gravity. The horizontal distance is then $v_x \cdot t$, where $v_x = v \cos \theta$ and $t$ is the time of flight. Given the complexity of the calculation and the need to account for the initial height, the answer is not straightforwardly derived from the basic projectile motion formula, but through a more detailed calculation that considers the initial height, the result is approximately 74.4 ft, making option B correct.

Solve the Problem 39∘39 ^ { \circ }39∘ To the Horizontal From

Solve the Problem $39 ^ { \circ }$ To the Horizontal From