If you can throw a ball vertically upward to a height h = 30.5 m, what is the maximum horizontal range over which you can throw the same ball, assuming you throw it at the same initial speed?
To find the maximum horizontal range over which you can throw the ball, we can use the concept of projectile motion.
When a ball is thrown vertically upward, its initial velocity has two components: the horizontal component and the vertical component. The vertical component determines the maximum height reached, while the horizontal component determines the range.
The vertical motion of the ball can be described by the equations:
h = (v_i^2 sin^2θ)/(2g) -- (1)
where h is the maximum height (given as 30.5 m), v_i is the initial velocity, θ is the angle of projection (90 degrees for vertical) and g is the acceleration due to gravity (approximately 9.8 m/s^2).
From equation (1), we can solve for the initial velocity v_i:
v_i = √(2gh) / sinθ -- (2)
Since the angle of projection for throwing a ball vertically is 90 degrees, the sinθ term in equation (2) becomes 1, and we can simplify the equation as:
v_i = √(2gh) -- (3)
Now, to find the horizontal range, we can use the equation:
R = v_i * t -- (4)
where R is the range, v_i is the initial velocity (as calculated in equation 3), and t is the total time of flight.
In projectile motion, the time taken to reach the maximum height is equal to the time taken to return to the ground level, and the total time of flight is twice that time.
So, we can calculate the total time of flight using the equation:
t = 2 * (√(2h/g)) -- (5)
Substituting the value of t (as calculated from equation 5) into equation 4, we can determine the horizontal range R:
R = v_i * 2 * (√(2h/g)) -- (6)
Using the given height h = 30.5 m and the value of g = 9.8 m/s^2, we can finally calculate the maximum horizontal range R by plugging them into equation 6.