Question

A basketball is tossed upwards with a speed of 5.0\,\dfrac{\text m}{\text s}5.0
s
m
​
5, point, 0, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction. We can ignore air resistance.
What is the maximum height reached by the basketball?

Answer 1

Answer:

Hey,

QUESTION)

• v0 = 5 m/s

The ball is only subject to its own weight, so according to Newton's first law, there is :

$\sum\overrightarrow{F_{ext}} = m \times \vec{a} \\ \vec{P} = m \times \vec{a} \\ m \times \vec{g} = m \times \vec{a} \\ \vec{g} = \vec{a} \\ \vec{a}\left(\begin{array}{c}a_x=0&a_y=-g\end{array}\right)$

$\text{Let's find the primitive of the acceleration vector}\\ \vec{v}\left(\begin{array}{c}v_x= v_0 \times sin(\theta) & v_y=-gt + v_0 \times sin(\theta)\end{array}\right)\\ \text{Let's find the primitive of the speed vector}\\ \overrightarrow{OM}\left(\begin{array}{c} x= v_0 \times sin(\theta) \times t & y=-\frac{1}{2} gt^2 + v_0 \times sin(\theta)\times t\end{array}\right)$

$\text{At maximum altitude, the y-axis velocity is zero such that}\\ v_y = 0 \Longleftrightarrow 0=-gt + v_0 \times sin(\theta) \Longleftrightarrow t = \frac{v_0\times sin(\theta)}{g} \ (1)\\ \text{So this time t corresponds to the moment when the ball t reaches the top.}$

Thus, knowing the theta value, we can find the maximum altitude by replacing t in the equation for y with the expression above.