Since two children with shoulder height h are playing catch in a gym with a ceiling height H, the maximum distance is d = √{8v₀²(H - h)/g - (H - h)²}
Since two children with shoulder height h are playing catch in a gym with a ceiling height H. The ball is thrown and caught at shoulder level (Figure 1). Children can throw the ball to each other at V0 speed. To find the maximum distance between children to play this game, we proceed as follows
We know that the horizontal distance d is the range of the ball (projectile).
So, d = v₀²sin2Ф/g where
- v = initial velocity of ball
- Ф = angle of projection and
- g = acceleration due to gravity
Now, the maximum height of the projectile is H - h. Also, the maximum height is given by
H - h = v₀²sin²Ф/2g
Using trigonometric identitiy cos2Ф = 1 - 2sin²Ф.
sin²Ф = (1 - cos2Ф)/2
Substituting this into the maximum height, we have that
H - h = v₀²sin²Ф/2g
H - h = v₀²(1 - cos2Ф)/2/2g
H - h = v₀²(1 - cos2Ф)/4g
H - h = (v₀² - v₀²cos2Ф)/4g
4g(H - h) = v₀² - v₀²cos2Ф
v₀²cos2Ф = v₀² - 4g(H - h) (1)
From d, we have that
v₀²sin2Ф = dg (2)
Squaring both equations (1) and (2) and adding, we have that
[v₀²cos2Ф]² = [v₀² - 4g(H - h)]²
+
(v₀²sin2Ф)² = (dg)²
[v₀²cos2Ф]² + (v₀²sin2Ф)² = [v₀² - 4g(H - h)]² + (dg)²
[v₀²}[(cos2Ф]² + (sin2Ф)²] = (v₀²)² - 8v₀²g(H - h) + [g(H - h)]² + (dg)²
[v₀²}² = (v₀²)² - 8v₀²g(H - h) + [g(H - h)]² + (dg)²
[v₀²}² - (v₀²)² = - 8v₀²g(H - h) + [g(H - h)]² + (dg)²
0 = - 8v₀²g(H - h) + [g(H - h)]² + (dg)²
-(dg)² = - 8v₀²g(H - h) + [g(H - h)]²
Dividing through by - 1, we have that
(dg)² = 8v₀²g(H - h) - [g(H - h)]²
Dividing through by g², we have that
(dg)²/g² = {8v₀²g(H - h) - [g(H - h)]²}/ g²
d² = {8v₀²(H - h)/g - (H - h)²}
Taking square root of both sides, we that
d = √{8v₀²(H - h)/g - (H - h)²}
So, the maximum distance is d = √{8v₀²(H - h)/g - (H - h)²}