A projectile is launched from and returns to ground level, as the figure shows.

ID: 1372345 • Letter: A

Question

There is no air resistance. The horizontal range of the projectile is measured to be

R = 250 m, and the horizontal component of the launch velocity is v0x = +38 m/s.

(i) What is the final value of the horizontal component vx of the projectile's velocity?

(ii) Can the time be determined for the horizontal part of the motion?

(iii) Is the time for the horizontal part of the motion the same as that for the vertical part?

(iv) For the vertical part of the motion, what is the displacement of the projectile? (in meters)

Thank you!!!

Here ,

i) as the horizontal velocity is same thorugh all the motion

final horizontal component vx = 38 m/s

ii)

yes , the time for the horizontal part of the motion can be determined.

iii)

yes . the time is sane for horizontal and vertical motion

iv)

time , t = 250/38

t = 6.579 s

Now ,for the vertical dispalcement = 0.5 * 9.8 * (6.579/2)^2

vertical dispalcement = 53.02 m

the vertical dispalcement is 53.02 m

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