Question

if the ball leaves the projectile launcher at a speed of 4.00 m/s at an angle of 30ᴼ, and the projectile launcher is on a table at a height of 1.2 m, how far horizontally will it hit the ground?

Answer 1

The ball will travel a horizontal distance of 0 meters before hitting the ground immediately after being launched.

Step-by-step explanation:

To find the distance the ball will travel horizontally, we need to analyze the horizontal and vertical components of its motion separately. Since the initial velocity of the ball is 4.00 m/s and it is launched at an angle of 30°, the horizontal component of its velocity can be found using the equation:

Vx = V * cos(θ)

where V is the initial velocity and θ is the launch angle. Plugging in the values, we have:

Vx = 4.00 m/s * cos(30°) = 4.00 m/s * 0.866 = 3.464 m/s

The time it takes for the ball to hit the ground can be found using the vertical motion equation:

y = Vy * t + 0.5 * g * t^2

Where y is the initial height (1.2 m), Vy is the vertical component of initial velocity, t is the time, and g is the acceleration due to gravity (-9.8 m/s^2). Set y equal to 0 and solve for t:

0 = Vy * t + 0.5 * g * t^2

Since the ball is initially launched horizontally, Vy is 0. Plugging in the values, we have:

0 = 0 + 0.5 * (-9.8 m/s^2) * t^2

simplifying the equation gives:

t = sqrt(0 / (-4.9 m/s^2)) = 0

Since the time is 0, the ball hits the ground immediately after it is launched. Therefore, the horizontal distance it travels is:

Distance = Vx * t

Plugging in the values, we have:

Distance = 3.464 m/s * 0 s = 0 m