Question

A basketball is tossed upwards with a speed of 5.0\,\dfrac{\text m}{\text s}5.0

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.