A baseball hit just above the ground leaves the bat at 27 m/s at 45∘ above the horizontal.
How far away does the ball strike the ground?
How much time is the ball in the air?
Vi = 27 sin 45
u = 27 cos 45
h = Vi t - 4.9 t^2
at ground h = 0
0 = Vi t -4.9 t^2 = t (Vi-4.9t)
t = Vi/4.9
then
range = u t
To solve this problem, we can use the equations of motion for projectiles.
The horizontal and vertical motions of the baseball are independent of each other. Let's break down the given information:
Initial speed (velocity) of the baseball, v₀ = 27 m/s
Launch angle, θ = 45 degrees
Acceleration due to gravity, g = 9.8 m/s² (assuming no air resistance)
First, let's find the time it takes for the ball to reach the highest point of its trajectory. We'll call this time t₀.
Using the equation of motion for the vertical component of motion:
v = v₀sin(θ) - gt
We plug in the values we have:
0 m/s = 27 m/s * sin(45) - 9.8 m/s² * t₀
Simplifying the equation, we get:
t₀ = (27 m/s * sin(45)) / (9.8 m/s²)
t₀ ≈ 1.85 seconds
Since the total time of flight is twice the time to reach the maximum height, the total time in the air (t) can be found by multiplying t₀ by 2:
t ≈ 1.85 seconds * 2 = 3.70 seconds
Now, let's find how far away the baseball strikes the ground. We'll call this distance d.
Using the equation of motion for the horizontal component of motion:
d = v₀ * cos(θ) * t
Plugging in the values we have:
d = 27 m/s * cos(45) * 3.70 seconds
Simplifying the equation, we get:
d ≈ 68.74 meters
Therefore, the baseball strikes the ground approximately 68.74 meters away, and it is in the air for about 3.70 seconds.