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.
(Vo^2/g)*sin(2*45) = 40.8 meters
(since it is caught at the same level it was thrown).
To determine how far the softball would travel, we can break down the motion into horizontal and vertical components.
First, let's calculate the time it takes for the softball to hit the ground using its vertical motion. We can use the equation:
h = ut + (1/2)gt^2
Where h is the initial height (2m), u is the initial vertical velocity, g is the acceleration due to gravity, and t is the time.
In this case, the initial vertical velocity is the vertical component of the initial velocity, which can be calculated using the equation:
u = v * sin(angle)
Where v is the initial velocity (20 m/s) and angle is the launch angle (45 degrees). Therefore, the initial vertical velocity is:
u = 20 m/s * sin(45 degrees)
Next, we can plug these values into the equation for height:
2m = (20 m/s * sin(45 degrees)) * t + (1/2)(9.8 m/s^2)(t^2)
Simplifying this equation, we can solve for t:
(1/2)(9.8 m/s^2)t^2 + (20 m/s * sin(45 degrees)) * t - 2m = 0
Solving this quadratic equation would give us the time (t) it takes for the softball to hit the ground.
Once we have the time, we can determine the horizontal distance traveled by multiplying the horizontal component of the initial velocity by the time:
d = u * cos(angle) * t
Where u is the initial horizontal velocity, which can be calculated using the equation:
u = v * cos(angle)
Plugging in the values, we have:
d = (20 m/s * cos(45 degrees)) * t
Calculating this equation will give us the distance (d) traveled by the softball.