Question

A Jodgraball is punted at 25.0 m/s [40.0°) on Bhavya's home planet. What is the range of the object on level ground? (Use g = 18.0 m/s²)

A) Calculate the range of the Jodgraball on level ground when punted at 25.0 m/s at an angle of 40.0°.
B) Determine the horizontal distance the Jodgraball travels on level ground after being punted at 25.0 m/s and an angle of 40.0°.
C) Use the given values to find the range of the Jodgraball when it's launched on Bhavya's home planet at 25.0 m/s and 40.0°.
D) Calculate the horizontal displacement of the Jodgraball after being punted at 25.0 m/s and 40.0° on Bhavya's home planet with a gravitational acceleration of 18.0 m/s².

Answer 1

The range of the Jodgraball on level ground when punted at 25.0 m/s at an angle of 40.0° is 49.31 m.

To calculate the range of the Jodgraball on level ground, we need to find the horizontal distance it travels. We can use the equation for horizontal distance (range) of a projectile:

R = (v^2 * sin(2θ)) / g

where v is the initial velocity of the projectile, θ is the launch angle, and g is the acceleration due to gravity. Plugging in the given values, we get:

R = (25.0^2 * sin(2 * 40.0)) / 18.0 = 49.31 m

Therefore, the range of the Jodgraball on level ground is 49.31 m.