with what velocity must a ball be thrown in order to achieve the height of 50 ft.? and how long will it stay in the air?
Equations of motion
v=u+at
v^2=u^2+2aS
S=ut+(1/2)*at^2
u = initial velocity
v = final velocity
t = time
a = acceleration
S = displacement.
Once you know 3 out of the 5 you can find the other two. See if you can write down 3 things you know about the situation.
To calculate the velocity required to achieve a specific height, we can use the equations of projectile motion. In this case, we'll use the equation for vertical displacement, which is given by:
h = (v^2 * sin^2(θ)) / (2 * g)
Where:
- h is the height (50 ft in this case)
- v is the initial velocity of the ball
- θ is the launch angle (assumed to be 45 degrees for simplicity)
- g is the acceleration due to gravity (32.2 ft/s^2)
By rearranging the equation, we can solve for the initial velocity (v):
v = sqrt((2 * g * h) / sin^2(θ))
Now, let's calculate the velocity required. Assuming the launch angle (θ) is 45 degrees:
θ = 45 degrees = π/4 radians
h = 50 ft
g = 32.2 ft/s^2
Plugging in these values into the equation:
v = sqrt((2 * 32.2 * 50) / sin^2(π/4))
Using the value of sin(π/4) = sqrt(2)/2:
v = sqrt((2 * 32.2 * 50) / (sqrt(2)/2)^2)
v = sqrt((2 * 32.2 * 50) / (2/4))
v = sqrt((2 * 32.2 * 50) / (1/2))
v = sqrt((2 * 32.2 * 50) / 0.5)
v = sqrt(2 * 32.2 * 50 * 2)
v ≈ 71.57 ft/s
Therefore, the ball must be thrown with a velocity of approximately 71.57 ft/s to reach a height of 50 ft.
To calculate how long the ball will stay in the air (time of flight), we can use the equation for vertical motion:
t = (2 * v * sin(θ)) / g
Using the same assumptions for θ and g:
t = (2 * 71.57 * (sqrt(2)/2)) / 32.2
t ≈ 2.00 seconds
Therefore, the ball will stay in the air for approximately 2.00 seconds.