Question

Problem Try to answer the following questions:(a) What is the maximum height above ground reached by the ball?(b) What are the magnitude and the direction of the velocity of the ball just before it hits the ground? Show Your Problem Solving Steps: Show these below:1) Draw a Sketch2) Choose origin, coordinate direction3) Inventory List – What is known?4) Write the kinematics equation(s) and solution of Part (a):5) Write the kinematics equation(s) and solution of Part (b):Problem 3 A small ball is launched at an angle of 30.0 degrees above the horizontal. It reaches a maximum height of 2.5 m with respect to the launch position. Find (a) the initial velocity of the ball when it’s launched and (b) its range, defined as the horizontal distance traveled until it returns to his original height. As always you can ignore air resistance.(a) Initial velocity [Hints: How is v0 related to vx0 and vy0. How can you use the information given to calculate either or both of the components of the initial velocity?](b) Range [Hints: This problem is very similar to today’s Lab Challenge except that for the challenge the ball will land at a different height.]

Answer 1

3.

`\begin{gathered} \theta=30^(\circ) \\ y_(\max )=2.5m \end{gathered}`

a)

`\begin{gathered} y_(\max )=(v^2\sin ^2(\theta))/(g) \\ \end{gathered}`

Solve for v:

`\begin{gathered} v=\sqrt[]{(y_(\max )\cdot g)/(\sin ^2(\theta))} \\ v=98\cdot(m)/(s) \end{gathered}`

b)

`\begin{gathered} r=(v^2)/(9)\sin (2\theta) \\ r=(98^2)/(9.8)\cdot\sin (2\cdot30) \\ r=(98^2)/(9.8)\sin (60) \\ r=848.7m \end{gathered}`