A ball launched from ground level lands 3 s later on a level field 48 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.)
To find the magnitude of the initial velocity vector and the angle above the horizontal, we can use the kinematic equations of motion.
Let's assume the initial velocity of the ball is V, the angle above the horizontal is θ, and the time of flight is t. We can break down the motion of the ball into horizontal and vertical components.
In the horizontal direction, there is no acceleration acting on the ball, so the velocity remains constant. The horizontal distance covered by the ball is given as 48 m, and the time of flight is 3 s. Therefore, we can use the formula:
Horizontal distance = Velocity × Time
48 m = V * cos(θ) * 3 s
Next, in the vertical direction, we can use the equations of motion to find the initial vertical velocity (Vy) and the time of flight (t).
The formula for vertical displacement (Δy) in terms of initial velocity, time, and acceleration is:
Δy = Vy * t + (1/2) * g * t^2
Since the ball lands at the same height it was launched (ground level), the vertical displacement is zero. So we have:
0 = Vy * 3s - (1/2) * g * (3s)^2
Simplifying this equation, we get:
2 * Vy = 9.8 m/s^2 * 9 s
Now we can substitute the value of Vy into the previous equation for horizontal distance:
48 m = V * cos(θ) * 3 s
Solving for V, we have:
V = 48 m / (3 s * cos(θ))
Now we can substitute the value of Vy into the equation for V:
V = 48 m / (3 s * (48 m / (3 s * cos(θ))))
Simplifying further, we get:
V = 16 m / cos(θ)
Therefore, the magnitude of the initial velocity vector is 16 m/s.
To find the angle θ, we can use the equation Vy = V * sin(θ):
2 * Vy = 9.8 m/s^2 * 9 s
2 * V * sin(θ) = 9.8 m/s^2 * 9 s
Substituting the value of V, we have:
2 * 16 m/s * sin(θ) = 9.8 m/s^2 * 9 s
Simplifying further:
sin(θ) = (9.8 m/s^2 * 9 s) / (2 * 16 m/s)
θ = arcsin((9.8 m/s^2 * 9 s) / (2 * 16 m/s))
Using a calculator, we find:
θ ≈ 43.6 degrees
Therefore, the magnitude of the initial velocity vector is approximately 16 m/s, and the angle above the horizontal is approximately 43.6 degrees.