A ball is kicked at an angle of 50 with a velocity of 14m/s.
a. how long is it in the air?
b. how high does it rise?
can someone please explain to me how to solve this???
horizontal velocity, U = 14 cos 50 = 9
Since there is no horizontal force on the ball, the horizontal velocity will remain 9.
Initial vertical velocity, Vo = 14 sin 50 = 10.7
vertical velocity at the top = 0 (it stops and starts down)
v = Vo - 9.8 t
0 = 10.7 - 9.8 t
t = 1.09 seconds to the top
It will spend the same time falling so a total of 2.18 seconds in the air.
now how high at 1.09 seconds?
h = 0 + Vo t -(1/2)9.8 t^2
h = 10.7 (1.09) - 4.9 * 1.19
h = 5.83 meters (about 17 feet)
thanks! I have question though...what happened to the ^2 in the second part? Its in the second step but then where does it go?
nevermind, i figured it out
To solve this problem, you can use the equations of motion for a projectile in two dimensions. Let's break it down step by step:
Step 1: Resolve the velocity into its horizontal and vertical components.
Given that the ball is kicked at an angle of 50 degrees with a velocity of 14 m/s, you can find the horizontal and vertical components of the velocity using trigonometry.
Horizontal component (Vx) = velocity * cos(angle)
Vertical component (Vy) = velocity * sin(angle)
Vy = 14 m/s * sin(50°)
Vy = 14 m/s * 0.766
Vy ≈ 10.724 m/s
Vx = 14 m/s * cos(50°)
Vx = 14 m/s * 0.64
Vx ≈ 8.96 m/s
Step 2: Determine the time of flight (how long the ball is in the air).
The time of flight is the total time it takes for the ball to go up and then come back down to the same level. We'll use the vertical component of the velocity to calculate this.
Use the equation: Time = 2 * Vy / acceleration due to gravity (g)
Acceleration due to gravity (g) is approximately 9.8 m/s^2.
Time = 2 * 10.724 m/s / 9.8 m/s^2
Time ≈ 2.194 s
Step 3: Calculate the maximum height reached by the ball.
To calculate the maximum height, we will use the vertical component of the velocity and the time of flight.
Use the equation: Height = Vy^2 / (2 * g)
Height = (10.724 m/s)^2 / (2 * 9.8 m/s^2)
Height ≈ 5.799 m
Therefore, the solutions to the problem are:
a. The ball is in the air for approximately 2.194 seconds.
b. The ball reaches a maximum height of approximately 5.799 meters.