2)a-A projectile is shot on level ground with a horizontal velocityof 23 m/s and

Question

2)a-A projectile is shot on level ground with a horizontal velocityof 23 m/s and a vertical
velocity of 43 m/s . The acceleration of gravity is 9.8 m/s^2 .Find the travel time of the projectile ttotal. Answer in units ofs.

b-Find the range R. Answer in units of m.

c-Now consider a new situation, where the magnitude of v0 is fixedto be the same value as that in part 1 and part 2, and we vary theangle to get the maximum range.
Find the height when the range is maximum. Answer in units ofm.

Thank you for your help!

Explanation / Answer

Calculate the time using the y velocity as if it were a freefall problem.
Vf = Vi + a(t) Vf = 0 (at the peak there is no acceleration) Vi = 43 m/s a = -9.8 m/s^2 t = ? 0 = 43 + -9.8(t) -43/-9.8 = t t = 4.39 s
0 = 43 + -9.8(t) -43/-9.8 = t t = 4.39 s
Now we want to find the maximum distance: delta x = initial velocity of x times time delta x = 23*4.39 delta x = 100.97
HINT the optimum angle for the farthest range is 45 degrees.

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