An object is launched with an initial velocity of 50.0 m/s at a launch angle of

Question

An object is launched with an initial velocity of 50.0 m/s at a launch angle of 36.9 above the horizontal.

PART A: Determine x-values at each 1 s from t = 0 s to t = 6 s (Express your answer using three significant figures. Give your answer in the time increasing order, separate them with commas.)

PART B: Determine y-values at each 1 s from t = 0 s to t = 6 s (Express your answer using three significant figures. Give your answer in the time increasing order, separate them with commas.)

PART C: Determine vx-values at each 1 s from t = 0 s to t = 6 s (Express your answer using three significant figures. Give your answer in the time increasing order, separate them with commas.)

PART D: Determine vy-values at each 1 s from t = 0 s to t = 6 s (Express your answer using three significant figures. Give your answer in the time increasing order, separate them with commas.)

PART E: Determine the speed v at each 1 s from t = 0 s to t = 6 s (Express your answer using three significant figures. Give your answer in the time increasing order, separate them with commas.)

PART F: Plot a graph of the object’s trajectory during the first 6 s of motion ( What would the graph look like)

Explanation / Answer

Part A:

Initial velocity of the projectile, U = 50.0 m/s at an angle 36.9 deh. from horizontal.

So, Ux = 50.0*cos36.9 = 40.0 m/s

Uy = 50.0*sin36.9 = 30.0 m/s

Horizontal velocity of the projectile shall be constant.

So, x(0) = 0

x(1) = 40.0 m, x(2) = 80.0 m, x(3) = 120.0 m, x(4) = 160.0 m, x(5) = 200.0 m, x(6) = 240.0 m

Part (B) -

For determining the y-component, the fravitational acceleration g will act vertically dowinward.

So, y(0) = 0

y(1) = 30*1 - 0.5*9.81*1^2 = 30 - 4.90 = 25.1 m

y(2) = 30*2 - 0.5*9.8*2^2 = 60 - 19.6 = 40.4 m

y(3) = 30*3 - 0.5*9.8*3^2 = 90 - 44.1 = 45.9 m

y(4) = 30*4 - 0.5*9.8*4^2 = 120 - 78.4 = 41.6 m

y(5) = 30*5 - 0.5*9.8*5^2 = 150 - 122.5 = 27.5 m

y(6) = 30*6 - 0.5*9.8*6^2 = 180 - 176.4 = 3.6 m

Part (C) :

Horizontal velocity shall be constant.

Means, at t=1 to 6

vx = 40.0 m/s

Part (D):

vy(1) = 30 - 9.8*1 = 20.2 m/s

vy(2) = 30 - 9.8*2 = 10.4 m/s

vy(3) = 30 - 9.8*3 = 0.6 m/s

other values of vy is negative means towards -ve y-axis. which can be assumed to 0.

Part E:

v = sqrt[vx^2 + vy^2]

put the values -

like at t = 1

v(1) = sqrt[40^2 + 20.2^2] = 44.8 m/s

like-wise you can calculate the values of v for t = 2 to 6

Part F:

Plot yourself the graph for the value of x and y for t = 0 to 6 s.Its very easy.

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