An object is launched with an initial velocity of 50.0 at a launch angle of 36.9

Question

An object is launched with an initial velocity of 50.0 at a launch angle of 36.9 above the horizontal.
A)Determine -values at each 1 from = 0 to = 6
Express your answer using three significant figures. Give your answer in the time increasing order, separate them with commas.
B)Determine -values at each 1 from = 0 to = 6
Express your answers using one decimal place. Give your answer in the time increasing order, separate them with commas.
C)Determine the speed at each 1 from = 0 to = 6
Give your answers in the time increasing order, separate them with commas.

Explanation / Answer

The function for the traveled distance is: s = s0 v.t + 1/2.a.t² for the velocity the formula is: v = v0 + a.t We consider both directions horizontal (x) and vertical (y) separate. In the horizontal (x) direction; x = vx0.t + 1/2.ax.t² vx = vx0 +ax.t In the x direction no acceleration is present. ax = 0 m/s² The velocity at start in the x-direction vx0 = 50.0 * cos(34) = 43.0 m/s so: x = 43.0.t vx = 43.0 In the vertical (y) direction: y = vy0.t + 1/2.ay.t² vy = vy0 + ay.t In the y direction the gravitation works: ay = -9.81 m/s² The velocity at start in the y-direction vy0 = 50.0 * sin(34) = 25.5 m/s so: vy = 25.5 - 9.81.t y = 25.5.t - 4.90.t² t=1 x=43.0 vx=43.0 y=30.4 vy=15.7 t=2 x=86.0 vx=43.0 y=31.4 vy=5.9 t=3 x=129.0 vx=43.0 y=32.4 vy=-3.9 (rockets goes down) t=4 x=172.0 vx=43.0 y=23.6 vy=-13.7 t=5 x=215 vx=43.0 y=5.0 vy =-23.6 t=6 x=258 vx=43.0 y=0 vy =0 (rocket has hit the ground) y=0 at 0=25.5.t - 4.90.t² t = 25.5/4.90 = 5.2 s

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