A ball launched from ground level lands 2.7 s later on a level field 50 m away from the launch point. Find the magnitude of the initial velocity vector and the angle it is above the horizontal. (Ignore any effects due to air resistance.)
A. Find the magnitude m/s
B. Find the direction ° (above the horizontal)
To find the magnitude of the initial velocity vector, we can use the horizontal distance traveled by the ball and the time it takes to reach that distance.
Given:
- Horizontal distance traveled (range): 50 m
- Time taken: 2.7 s
We know that the horizontal component of the initial velocity remains constant throughout the motion since there is no force acting on the ball in the horizontal direction.
1. Calculate the horizontal component of the initial velocity (Vx):
Vx = (horizontal distance traveled) / (time taken)
Vx = 50 m / 2.7 s
Vx ≈ 18.5185 m/s
Now, let's find the vertical component of the initial velocity. Since the ball is launched from ground level and lands on a level field, the displacement in the vertical direction is zero. We can use the equation of motion to find the vertical component of the initial velocity.
2. Calculate the vertical component of the initial velocity (Vy):
Displacement (d) = 0 (since the ball lands at the same height above the ground)
Initial velocity (Viy) = ?
Time taken (t) = 2.7 s
Acceleration due to gravity (g) = 9.8 m/s^2
d = Viy * t - (1/2) * g * t^2
Since d = 0, we can rewrite the equation as:
0 = Viy * t - (1/2) * g * t^2
Rearranging the equation:
Viy * t = (1/2) * g * t^2
Viy = (1/2) * g * t
Viy = (1/2) * 9.8 m/s^2 * 2.7 s
Viy ≈ 13.235 m/s
Now we have both the horizontal and vertical components of the initial velocity. To find the magnitude and angle of the initial velocity vector, we can use the Pythagorean theorem and trigonometric calculations.
3. Calculate the magnitude of the initial velocity:
Magnitude (V) = √(Vx^2 + Vy^2)
V = √((18.5185 m/s)^2 + (13.235 m/s)^2)
V ≈ √(343.186225 + 175.050225)
V ≈ √(518.23645)
V ≈ 22.7888 m/s
4. Calculate the angle above the horizontal:
Angle (θ) = tan^(-1)(Vy / Vx)
θ = tan^(-1)((13.235 m/s) / (18.5185 m/s))
θ ≈ 35.355°
Therefore, the magnitude of the initial velocity vector is approximately 22.7888 m/s, and the angle above the horizontal is approximately 35.355°.
For a projectile fired at an angle of µº to the horizontal with a velocity of Vm/s, the maximum height reached derives from h = V^2(sin^2(µ))/29 and the distance traveled parallel to the ground derives from d = V^2(sin(2µ))/g
1--Vcos(µ) = 50/2.75 = 18.181818
2--Vsin(µ) = 9.8(2.75) = 26.95
3--Vsin(µ)/Vcos(µ) = 26.95/18.18 = 1.48239
4--(µ) = arctan(1.48239) = 56º
5--Therefore, the initial launch velocity is V = 18.1818/cos(56) = 32.51m/s in the direction of 56º above the horizontal.
I hope this has been of some help to you.