A ball is launched from ground level at 31.8 m/s at an angle of 35° above the horizontal. How far does it go before it is at ground level again?

A) 14 m
B) 21 m
C) 43 m
D) 86 m
E) 97 m

21

The vertical component of velocity determines how long the ball is in the air. Multiply that by the horizontal component to determine how far it goes before hitting the ground

How do you obtain the vertical and horizontal components? is that the 31.8sin35 and 31.8cos35?

Thanks

Yes

To find the distance the ball travels before it is at ground level again, we can use the equations of motion for projectile motion.

First, we need to break the initial velocity into its horizontal and vertical components. The horizontal component (Vx) is given by V * cos(θ), where V is the magnitude of the velocity (31.8 m/s) and θ is the angle (35°). Therefore, Vx = 31.8 m/s * cos(35°).

Next, we can calculate the time it takes for the ball to reach ground level again by considering the vertical motion. The time of flight can be calculated using the equation: t = 2 * Vy / g, where Vy is the vertical component of velocity and g is the acceleration due to gravity (approximately 9.8 m/s²). Vy is given by V * sin(θ), so Vy = 31.8 m/s * sin(35°).

Now, we can calculate the total horizontal distance traveled by the ball using the equation: d = Vx * t.
Substituting the values we found, d = (31.8 m/s * cos(35°)) * (2 * (31.8 m/s * sin(35°)) / 9.8 m/s²).

Calculating this expression gives us approximately 97 meters. Therefore, the ball travels a distance of approximately 97 meters before it is at ground level again. Therefore, the correct answer is (E) 97 m.