Question

show answer No Attempt Treated as a projectile, what is the maximum range, in meters, obtainable by a person if he or she has a take-off speed of 9.2 m/s? Assume the motion is over level ground and the initial velocity makes an angle of 45° with the horizontal.

Answer 1

Answer: R=8.64 meters

Explanation:

in the attachment

Answer 2

Answer:

8.637 m

Explanation:

Using

R = (U²sin2θ)/g.................. Equation 1

Where R = Range, U = initial velocity, θ = angle of projection, g = acceleration due to gravity.

Note: For maximum Range, θ = 45°, then sin2θ = sin90° = 1

Therefore,

Rmax = U²/g..................... Equation 2

Given: U = 9.2 m/s, g = 9.8 m/s²

Substitute into equation 2

Rmax = 9.2²/9.8

Rmax = 84.64/9.8

Rmax = 8.637 m

Hence the maximum range = 8.637 m