A projectile is fired at v 0 = 381.0 m/s at an angle of θ = 73.5 ∘ , with respect to the horizontal. Assume that air friction will shorten the range by 34.1 % . How far will the projectile travel in the horizontal direction, R ?

Explanation:

velocity of projection, vo = 381 m/s

angle of projection, θ = 73.5°

The formula for the range is

R = 8067.4 m

Range in shorten by 34.1 %

So, the new range is

R' = 8067.4 - 34.1 x 8067.4/100

R' = 5316.4 m

where's the following? you gotta add it to know what the choices are

explanation:

  distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

  slope of a graph = change in value of y -axis / change in values of x -axis

  comparing both the equations

    change in value of y -axis = change in position

  change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.