A volleyball is sent over a net, having been launched at a speed 5 m/s at an angle 42.7 degrees above the horizontal, from the ground. What is the ball’s speed, in m/s, 2.97 seconds after being launched?
To find the ball's speed 2.97 seconds after being launched, we need to break down the initial velocity of the ball into its horizontal and vertical components.
The horizontal component of the velocity remains constant throughout the motion, so we can use it to find the horizontal distance covered by the ball.
The horizontal component of the initial velocity can be found using the equation:
Vx = V * cos(θ)
where Vx is the horizontal component of velocity, V is the initial velocity, and θ is the launch angle.
Substituting the given values:
Vx = 5 m/s * cos(42.7°)
Next, we can find the horizontal distance covered by the ball using the equation:
horizontal distance = Vx * time
where time is given as 2.97 seconds.
Now, let's solve for the horizontal distance:
horizontal distance = (5 m/s * cos(42.7°)) * 2.97 s
Now, we can find the vertical component of velocity using the equation:
Vy = V * sin(θ)
Substituting the given values:
Vy = 5 m/s * sin(42.7°)
The vertical distance covered by the ball can be found using the equation:
vertical distance = (Vy * time) + (0.5 * g * time^2)
where g is the acceleration due to gravity on Earth, approximately 9.8 m/s^2.
Now, let's solve for the vertical distance:
vertical distance = (5 m/s * sin(42.7°)) * 2.97 s + (0.5 * 9.8 m/s^2 * (2.97 s)^2)
Finally, we can calculate the ball's speed 2.97 seconds after being launched using the Pythagorean theorem:
speed = √(horizontal distance^2 + vertical distance^2)
Plug in the values we calculated earlier to get the answer in meters per second (m/s).
vx = 5cos42.7 (it's constant)
vy = 5sin42.7 - gt
Then pythagorean for total velocity