a projectile is thrown from the ground with an initial velocity of 20.0m/s at an angle of 40 degrees above the horizontal, find the maximum height,the time required to reach its maximum height,c) it velocity at the top of the trajectory.d) the range of the projectile,e) the total time of the flight
Vo = 20m/s[40o]
Xo = 20*Cos40 = 15.32 m/s.
Yo = 20*sin40 = 12.86 m/s.
a. Y^2 = Yo^2 + 2g*h
Y = 0
Yo = 12.86 m/s.
g = -9.8 m/s^2
Solve for h.
b. Y = Yo + g*Tr
Y = 0
Yo = 12.86 m/s.
g = -9.8 m/s^2.
Solve for Tr.(Rise time or time to reach max. ht.)
c. V = Xo + Yi = Xo + 0 = Xo
d. Range = Vo^2*sin(2A)/g
Vo = 20 m/s
A = 40o
g = 9.8 m/s^2.
Solve for Range.
e. Tf = Tr = Fall time.
T = Tr+Tf = 2Tr = Time in flight.
To find the maximum height, time required to reach the maximum height, velocity at the top of the trajectory, range, and total time of flight of the projectile, we can use the following equations:
1. Maximum height (h):
We can use the kinematic equation for vertical motion:
h = (vi^2 * sin^2(θ)) / (2 * g)
where:
- h is the maximum height,
- vi is the initial velocity (20.0 m/s),
- θ is the angle of projection (40 degrees),
- g is the acceleration due to gravity (9.8 m/s^2).
1. Calculate h:
h = (20^2 * sin^2(40)) / (2 * 9.8)
h ≈ 12.7 meters
So, the maximum height is approximately 12.7 meters.
2. Time required to reach maximum height (t):
Again, using the kinematic equation for vertical motion, we can find the time it takes to reach the maximum height:
t = (vi * sin(θ)) / g
2. Calculate t:
t = (20 * sin(40)) / 9.8
t ≈ 1.29 seconds
So, the time required to reach the maximum height is approximately 1.29 seconds.
3. Velocity at the top of the trajectory (v):
We can find the vertical component of the velocity using:
v = vi * sin(θ)
3. Calculate v:
v = 20 * sin(40)
v ≈ 12.9 m/s
So, the velocity at the top of the trajectory is approximately 12.9 m/s.
4. Range (R):
The range of the projectile is the horizontal distance covered. We can use the formula:
R = vi * cos(θ) * t
4. Calculate R:
R = 20 * cos(40) * (2 * 1.29)
R ≈ 38.7 meters
So, the range of the projectile is approximately 38.7 meters.
5. Total time of flight (T):
The total time of flight is equal to twice the time to reach the maximum height (t):
T = 2 * t
5. Calculate T:
T = 2 * 1.29
T ≈ 2.58 seconds
So, the total time of flight is approximately 2.58 seconds.