A Softball is Thrown at an angle of 45 degrees with the horizontal. It leaves the hand 2m above the ground. Its Initial Velocity is 20 m/s . Neglecting all external resistance, how far would it travel ? The point at which it hits the ground is at the same level as the ground at the point from which it was thrown.
The vertical component of the initial velocity is 20/√2 = 14.14 m/s
So, the height
y = 2 + 14.14t - 4.9t^2
when does y=0?
At t=3.02 sec
Since the horizontal component of the velocity does not change, in those 3.02 seconds, the ball traveled
3.02*14.14 = 42.7 m
why the initial velocity u divided 20/√2. i cant understand. can u explain it ..
sin 45 = 1/√2
To find the horizontal distance the softball will travel, we need to analyze its motion and break it into its horizontal and vertical components.
Given:
Initial velocity (v₀) = 20 m/s
Angle of projection (θ) = 45 degrees
Initial height (h) = 2 m
First, let's find the time it takes for the softball to hit the ground. Since we know the initial velocity and angle of projection, we can use the equation for the vertical component of motion:
v_vert = v₀ * sin(θ)
where v_vert is the vertical component of the initial velocity.
v_vert = 20 m/s * sin(45 degrees) = 20 m/s * √2 / 2 = 10√2 m/s
The vertical motion can be divided into two parts: the time taken to reach the maximum height and the time taken to reach the ground. At the maximum height, the vertical component of velocity becomes zero.
Using the equation v_vert = at, where 'a' is the acceleration due to gravity (approximately 9.8 m/s²), we can find the time taken to reach the maximum height:
0 = 10√2 m/s - 9.8 m/s² * t_max
t_max = 10√2 m/s / 9.8 m/s²
t_max ≈ 1.44 s
Since the softball is thrown from a height of 2 m, it takes the same time (1.44 s) to reach the maximum height as it does to fall back to the ground. Therefore, the total time of flight is:
t_total = 2 * t_max
t_total ≈ 2 * 1.44 s ≈ 2.88 s
Now, let's find the horizontal distance traveled by the softball. We can use the equation for the horizontal component of motion:
v_horiz = v₀ * cos(θ)
where v_horiz is the horizontal component of the initial velocity.
v_horiz = 20 m/s * cos(45 degrees) = 20 m/s * √2 / 2 = 10√2 m/s
The horizontal distance (d) is given by:
d = v_horiz * t_total
= 10√2 m/s * 2.88 s
≈ 28.3 m
Therefore, the softball would travel approximately 28.3 meters horizontally before hitting the ground, neglecting all external resistance.