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

Answer 1

Given:
v = 50.0 m/s, the launch velocity
θ = 36.9°, the launch angle above the horizontal

Assume g = 9.8 m/s² and ignore air resistance.
The vertical component of the launch velocity is
Vy = (50 m/s)*sin(50°) = 30.02 m/s

The time, t, to reach maximum height is given by
(30.02 m/s) - (9.8 m/s²)*(t s) = 0
t = 3.0634 s
The time fo flight is 2*t = 6.1268 s

The horizontal velocity is
u = (50 m/s)cos(36.9°) = 39.9842 m/s
The horizontal distance traveled at time t is given in the table below.

Answer:

t, s x, m
------ --------
0 0
1 39.98
2 79.79
3 112.68
4 159.58
5 199.47
6 239.37

Answer 2

Explanation :

It is given that :

Initial velocity,
`v_0=50\ m/s`

Launching angle,
`\theta_0=36.9^0`

The motion followed by this object is called as its projectile motion.

the horizontal component is given by:

`x_t=x_0+v_0cos\theta_0t`.......(1)

Put all values in equation (1)

For t = 0,
`x_t=0`

For t = 1 s,
`x_1=0+50\ cos(36.9)\ 1=39.95\ m`

For t = 2 s,
`x_2=0+50\ cos(36.9)\ 2=79.9\ m`

For t = 3 s,
`x_3=0+50\ cos(36.9)\ 3=119.85\ m`

For t = 4 s,
`x_3=0+50\ cos(36.9)\ 4=159.8\ m`

For t = 5 s,
`x_4=0+50\ cos(36.9)\ 5=199.75\ m`

For t = 6 s,
`x_3=0+50\ cos(36.9)\ 6=239.7\ m`

Hence, this is the required solution.