Question

Find the horizontal range of a projectile launched at an angle of 40.0 ∘ with the initial velocities below. What is the range of the projectile with an initial speed of 33.0 m/s ?

Answer 1

The horizontal range of a projectile launched at an angle of 40.0 degrees with an initial speed of 33.0 m/s would be approximately 109.25 meters.

Step-by-step explanation:

To find the horizontal range of a projectile launched at an angle, we can use the following formula:

R = v_0^2 sin(2theta) / g

Where:

• R is the range
• v_0 is the initial velocity
• theta is the launch angle
• g is the acceleration due to gravity (approx. 9.81 m/s^2)

Given that the initial speed v_0 is 33.0 m/s and the launch angle theta is 40.0 degrees:

R = 33.0^2 sin(2 . 40.0) / 9.81

Now, compute the sine part:

sin(2 . 40.0) = sin(80.0) = 0.9848 (approx)

Now, calculate R:

R = {33.0^2 . 0.9848} / {9.81} = {1089 . 0.9848} / {9.81} = 109.25 (approx)

Therefore, the horizontal range of the projectile would be approximately 109.25 meters.