A projectile motion thrown with velocity of 200m/s direction of 53 degree above the horizontal?determine height,range,total time,time at particle reach maximum height?
To solve this problem, we can break down the projectile motion into two components: horizontal and vertical.
1. Determine the Vertical Component:
The initial vertical velocity (Viy) can be found using the given initial velocity (Vi) and the given angle above the horizontal (θ). We can use trigonometry to find the vertical component:
Viy = Vi * sin(θ)
Substituting the given values:
Viy = 200 m/s * sin(53°)
2. Determine the Horizontal Component:
The initial horizontal velocity (Vix) can also be found using the given initial velocity (Vi) and the given angle above the horizontal (θ). We can again use trigonometry to find the horizontal component:
Vix = Vi * cos(θ)
Substituting the given values:
Vix = 200 m/s * cos(53°)
3. Determine the time of flight (total time):
The total time of flight (T) can be found by dividing the vertical component (Viy) by the acceleration due to gravity (g). The acceleration due to gravity is approximately 9.8 m/s².
T = (2 * Viy) / g
Substituting the given values:
T = (2 * 200 m/s * sin(53°)) / 9.8 m/s²
4. Determine the range:
The range (R) can be found by multiplying the horizontal component (Vix) by the time of flight (T).
R = Vix * T
Substituting the given values:
R = (200 m/s * cos(53°)) * [(2 * 200 m/s * sin(53°)) / 9.8 m/s²]
5. Determine the maximum height:
The maximum height (H) can be found using the formula:
H = (Viy^2) / (2 * g)
Substituting the given values:
H = [200 m/s * sin(53°)]^2 / (2 * 9.8 m/s²)
6. Determine the time at which the particle reaches maximum height:
The time at which the particle reaches maximum height can be found by dividing the vertical component (Viy) by the acceleration due to gravity (g).
t_max_height = Viy / g
Substituting the given values:
t_max_height = (200 m/s * sin(53°)) / 9.8 m/s²
By substituting the given values in the equations above, you can calculate the height, range, total time, and time at which the particle reaches maximum height.
solve two problems separately
Problem 1 - vertical
Vi = 200 sin 53
v = Vi - 9.8 t
top when
0 = Vi -9.8 t solve for t
That t is half the time it is in the air
h = Vi t - 4.9 t^2 at time t
total time in air = 2 t
Problem 2, horizontal
u = Vi cos 53 until it crashes
d = u * total time = 2 u t
Max Height = 1,306M
Range = 3,840M
Hangtime/Total Time = 32 seconds
Time to reach max height = 16 seconds