Question

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

Determine x-values at each 1 s from t = 0 s to t = 6 s.

Determine y-values at each 1 s from t = 0 s to t = 6 s.

Determine vx-values at each 1 s from t = 0 s to t = 6 s.

Determine vy-values at each 1 s from t = 0 s to t = 6 s.

Determine the speed v at each 1 s from t = 0 s to t = 6 s.

Plot a graph of the object’s trajectory during the first 6 s of motion.

Answer 1

along X - direction

V_ox = 50 cos36.9 = 40 m/sec

so now at t = 0 , X = 0

at t = 1 , X = V_ox t = 40 X 1 = 40

at t = 2, X = V_ox t = 40 X 2 = 80

at t = 3 , X = V_ox t = 40 X 3 = 120

at t = 4 , X = V_ox t = 40 X 4 = 160

at t = 5 , X = V_ox t = 40 X 5 = 200

at t = 6 , X = V_ox t = 40 X 6 = 240

along the Y - direction the values are

at t = 1 ,V_fy = V_oy + ( -9.8 ) t = 30 - 9.8 X 1 = 20.2

at t = 2 ,V_fy = V_oy + ( -9.8 ) t = 30 - 9.8 X 2 = 10.4

at t = 3 ,V_fy = V_oy + ( -9.8 ) t = 30 - 9.8 X 3 = 0.6

at t = 4 ,V_fy = V_oy + ( -9.8 ) t = 30 - 9.8 X 4 = - 9.2

at t = 5 ,V_fy = V_oy + ( -9.8 ) t = 30 - 9.8 X 5 = - 19

at t = 6 ,V_fy = V_oy + ( -9.8 ) t = 30 - 9.8 X 6 = - 28.8

along Y - direction the values are about acceleration

a t t = 1 , Y = V_oy +1/2 gt² = 30 X 1 - 4.9 X 1² = 25

a t t = 2 , Y = V_oy +1/2 gt² = 30 X 1 - 4.9 X 2² = 40.4

a t t = 3 , Y = V_oy +1/2 gt² = 30 X 1 - 4.9 X 3² = 46

a t t = 4 , Y = V_oy +1/2 gt² = 30 X 1 - 4.9 X 4² = 41.5

a t t = 5 , Y = V_oy +1/2 gt² = 30 X 1 - 4.9 X 5² = 27.5

a t t = 6 , Y = V_oy +1/2 gt² = 30 X 1 - 4.9 X 6² = 3.5