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 spring gun at ground level fires a tennis ball at an angle of $33 ^ { \circ }$. The ball lands $12 \mathrm {~m}$ away. What was the ball's initial speed? Round your answer to the nearest tenth.
A) 14.7 m/sec
B) 11.3 m/sec
C) 128.7 m/sec
D) 3.6 m/sec

Question

Quizplus · Accepted Answer

Using the kinematic equations of motion, we can solve for the initial velocity of the tennis ball. Let's use the following variables:
- v0: initial velocity (what we're solving for)
- θ: launch angle (given as 33°)
- d: horizontal distance traveled (given as 12 m)
- g: acceleration due to gravity (9.8 m/s^2)

From the given information, we know that:
- The ball was launched from ground level, so its initial height (y0) is 0.
- The ball landed 12 m away horizontally, so its final horizontal position (xf) is 12 m.
- The ball hit the ground, so its final height (yf) is 0.

We can use the following kinematic equations to solve for v0:
- xf = v0*cos(θ)*t
- yf = v0*sin(θ)*t - (1/2)*g*t^2
- t = (xf) / (v0*cos(θ))

Substituting the given values and rearranging, we get:
- 12 = v0*cos(33°)*[(12) / (v0*cos(33°))]
- 0 = v0*sin(33°)*[(12) / (v0*cos(33°))] - (1/2)*g*[(12) / (v0*cos(33°))]^2

Simplifying and solving for v0, we get:
- v0 = 11.3 m/s

Therefore, the answer is B.

Solve the Problem 33∘33 ^ { \circ }33∘ the Ball Lands

Solve the Problem $33 ^ { \circ }$ the Ball Lands