Here is the question. I calculated a bunch of the data for you, and included it

Question

Here is the question. I calculated a bunch of the data for you, and included it with the question. This just a conformation I got it right.

A projectile is launched at an angle of 30 degrees above the horizontal. It is launched with a velocity of 58.3 m/s^2 (X velocity = 50.5, y velocity = 29.15), and travels 300m (horizontal distance). The total flight time is t=5.94s. At t=4.625s & t=1.32s the projectile will be 30m above the horizontal. How far (horizontally) from the endpoint will the projectile be at those times?

I believe that acceleration in the x direction should be -8.5. That being said the distance from the Stopping Point should be 157m & 241m. (59m & 143m from the starting point). Am i right? Please show work.

Explanation / Answer

there is no acceleration in horizontal.

so a = 0

we can find distance d = Vx * t

Vx = horizontal component of velocity ( this will be constant)

Vx = 58.3cos30 =50.5 m/s

at time t = 4.625 , Dx1 = 50.5 x 4.625 =233.51 m from starting point.

DIstance form end point = 300 - 233.51 = 66.50 ...........Ans

at time t = 1.32 sec , Dx2 = 50.5 x 1.32 = 66.66 m from starting point,

distance from end point = 233.34 m ........Ans

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects