Question

Ms. Leach picks up the watermelon and tries to throw it into a trashcan (assume the same height as her throwing arm). She throws it with a velocity of 12.2m/s at an angle of 530 measured up from the ground. How far away does it land?

Answer 1

the watermelon will land approximately 16.935 meters away from Ms. Leach.

Step-by-step explanation:

To calculate the horizontal distance the watermelon lands, we'll need to break down the initial velocity into its vertical and horizontal components. The horizontal component of the velocity can be determined by multiplying the initial velocity (12.2 m/s) by the cosine of the launch angle (53 degrees), while the vertical component can be found by multiplying the initial velocity by the sine of the launch angle.

Horizontal velocity (Vx) = initial velocity (12.2 m/s) * cosine(launch angle 53°)

Vertical velocity (Vy) = initial velocity (12.2 m/s) * sine(launch angle 53°)

Using these equations, we find:

Vx = 12.2 m/s * cos(53°) ≈ 8.8006 m/s

Vy = 12.2 m/s * sin(53°) ≈ 9.4294 m/s

Now, we can use the vertical velocity and gravitational acceleration (g = 9.8 m/s²) to determine the time taken (t) for the watermelon to reach its maximum height:

Vy = g * t

9.4294 m/s = 9.8 m/s² * t

t ≈ 0.9621 seconds

Since the total flight time will be twice the time to reach the maximum height (due to symmetry), the total time (T) can be calculated:

T = 2 * t

T ≈ 2 * 0.9621 s ≈ 1.9242 seconds

Using the horizontal velocity (Vx) and the total flight time (T), we can find the horizontal distance (D) traveled by the watermelon:

D = Vx * T

D ≈ 8.8006 m/s * 1.9242 s ≈ 16.935 meters