A projectile is shot on level ground with a
horizontal velocity of 21 m/s and a vertical
velocity of 45 m/s.
Part 1) Find the time the projectile is in the air.
The acceleration due to gravity is 9.8 m/s^2 .Answer in units of s.
Part 2) Find the range R. Answer in units of m.
Part 3) Let the magnitude of v0 be the same as in part 1, but consider the angle θ required to obtain the maximum range.
Find the height attained for the maximum range.Answer in units of m.
For my answer this what but it seem to be the wrong answer. Please help me.
1)
d = Viy t + 1/2 g t^2
d = 0
0 = 45 t - 4.9 t^2
4.9 t = 45
t = 9.2 s
2)R = X = Vx t = 21(9.2) = 193 m
3)Vo^2 = 21^2 + 45^2 = 246.6
Vo = 49.66 m/s or about 50
then at 45°
Vx = Vo sin45 = 49.66 sin45 = 35.1 or 35 m/s
d = Vyi t + 1/2 g t^2
0 = 35 t - 4.9 t^2
4.9 t = 35
t = 7.1 s
Xo = 21 m/s.
Yo = 45 m/s.
1. Y = Yo + g*Tr = 0 at max.ht.
Tr = -Yo/g = -45/-9.8 = 4.59 s. =
Rise time.
Tf = Tr = 4.59 s. = Fall time.
Tr+Tf = 4.59 + 4.59 = 9.18 s. = Time in air.
2. R = Xo*(Tr+Tf) = 21m/s * 9.18s
= 193 m.
3. A = 45o for max range.
Vo = Sqrt(21^1+45^2) = 49.7 m/s[45o].
Yo = 49.7*sin45 = 35.11 m/s.
Y^2 = Yo^2 + 2g*h = 0.
h = -(Yo^2)/2g = -(35.11)^2/-19.6
= 63 m.
Part 1) Time the projectile is in the air: Your calculation is correct. The projectile is in the air for 9.2 seconds.
Part 2) The range (horizontal distance) covered by the projectile: Your calculation is correct. The range is 193 meters.
Part 3) Height attained for the maximum range: Your calculation is incorrect. Let's try it again.
We know that the projectile reaches its maximum range when the angle of projection (θ) is 45 degrees. At this angle, the vertical velocity and horizontal velocity have the same magnitude.
So, the initial speed (magnitude of v0) is 50 m/s (not 49.66).
To find the height attained for the maximum range, we need to determine the time of flight when the projectile is at its maximum range. Since the projectile is in the air for 9.2 seconds, the time of flight for the maximum range will be half of that.
t_max_range = 9.2 / 2 = 4.6 seconds
Now, let's find the height attained at this time:
d = Vyi t + 1/2 g t^2
d = 45(4.6) - 4.9(4.6)^2
d = 206.1 - 4.9(21.16)
d = 206.1 - 103.684
d ≈ 102.416 meters
Therefore, the height attained for the maximum range is approximately 102.416 meters.
Your calculations for part 1 and part 2 are correct. The projectile is in the air for 9.2 seconds, and the range is 193 meters.
However, there seems to be an error in your calculation for part 3. The velocity at 45 degrees cannot be calculated by simply taking the square root of the sums of the squares of the horizontal and vertical velocities. It is necessary to use the trigonometric functions to find the correct value.
To find the maximum range, we need to find the angle at which the projectile should be launched. The maximum range is obtained when the launch angle is 45 degrees (or pi/4 radians), as this gives the maximum ratio of horizontal distance to vertical distance.
Using the given horizontal velocity (Vx = 21 m/s), the time of flight can be calculated as follows:
t = R / Vx = 193 / 21 = 9.19 s (rounded to two decimal places)
Now, using the time of flight, we can find the height attained for the maximum range. The height (h) can be calculated using the following formula:
h = V0y * t - (1/2) * g * t^2
Substituting the values:
h = 45 * 9.19 - (1/2) * 9.8 * (9.19)^2
h = 411.55 - 405.5866
h ≈ 5.97 m (rounded to two decimal places)
Therefore, the height attained for the maximum range is approximately 5.97 meters.
Based on the information you provided, let's go through each part of the problem and check the calculations:
Part 1) Find the time the projectile is in the air.
To find the time the projectile is in the air, you correctly set up the equation using the kinematic equation:
d = Viy t + 1/2 g t^2 (equation 1)
where d is the vertical displacement, Viy is the vertical component of the initial velocity, g is the acceleration due to gravity, and t is the time.
You correctly set d = 0 since the projectile lands on level ground. Then, you substituted Viy = 45 m/s and g = 9.8 m/s^2 into equation 1.
However, when solving for t, it seems you made an arithmetic error - the correct equation should be:
4.9 t^2 - 45 t = 0
This equation can be factored as:
t(4.9t - 45) = 0
which gives two possible solutions:
t = 0 (which is not applicable in this case since it represents the starting time) or
4.9t - 45 = 0
Solving this equation, we find:
4.9t = 45
t = 45/4.9
t ≈ 9.18 s (rounded to two decimal places)
So, the correct answer for part 1) is approximately 9.18 seconds.
Part 2) Find the range R.
The range or horizontal displacement, R, is given by the equation:
R = Vx * t (equation 2)
where Vx is the horizontal component of the initial velocity, and t is the time.
You correctly calculated Vx = 21 m/s and found t ≈ 9.18 s in part 1.
Substituting these values into equation 2, we get:
R = 21 * 9.18
R ≈ 195.18 m (rounded to two decimal places)
So, the correct answer for part 2) is approximately 195.18 meters.
Part 3) Find the height attained for the maximum range.
To find the height attained for the maximum range, we need to determine the launch angle that gives the maximum range. This occurs when the launch angle θ is 45°.
You correctly calculated the magnitude of the initial velocity, Vo, as:
Vo = √(21^2 + 45^2)
Vo ≈ 49.66 m/s (rounded to two decimal places)
Then, by considering the launch angle θ = 45°, you calculated the horizontal component of the initial velocity, Vx, as:
Vx = Vo * sin(45°)
Vx ≈ 49.66 * sin(45°)
Vx ≈ 35.15 m/s (rounded to two decimal places)
Next, you set up the equation for the vertical displacement, d, using equation 1 and substituted Vx = 35 m/s:
0 = 35 t - 4.9 t^2
However, the calculations for this part of the problem are incomplete. To find the height attained for the maximum range, you need to solve this equation for t and substitute the values back into equation 1 to find the corresponding displacement.
I hope this explanation helps clarify your solution and guide you in completing the calculation for part 3) of the problem.