Recall the simplified equation for projectile motion:[br][center][br][math]y=-\frac{g}{v^2}x^2+x+y_0,[/math][br][/center]where [math]v\ \text{m}/\text{s}[/math] denote the initial speed, [math]y_0\ \text{m}[/math] denotes the initial height, and [math]g \approx 9.81\ \text{m}/\text{s}^2[/math] denotes the [b]gravitational constant[/b] on earth.
You are standing [math]5\ \text{m}[/math] away from the basket, which is [math]3\ \text{m}[/math] above the ground. Suppose you are [math]1.5\ \text{m[/math] tall and release the ball at this height.[br][br]What speed [math]v\ \text{m}/\text{s}[/math] should you release the ball at?
Suppose instead you are [math]1.75\ \text{m}[/math]. At what speed should you throw the ball?
[math]8.1\ \text{m}/\text{s}[/math]
At this speed, what is the maximum height that your ball reaches?
[math](3.4 \pm 0.05)\ \text{m}[/math]
Find the maximum height in terms of [math]v[/math], [math]g[/math], and [math]y_0[/math].
[math]\frac{v^2}{4g}+y_0[/math]
Suppose the basket is [math]5\ \text{m}[/math] above the ground. Is it possible for you to throw the ball into the basket? Why or why not?