Question

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

To find the initial velocity, use the equation v0 = vy0/sin(theta), and to find the range use the equation R = (v0^2sin(2theta))/g.

Step-by-step explanation:

To find the initial velocity of the ball, we can use the fact that the ball reaches its maximum height of 2.5 m. The maximum height can be found using the equation:

h = (vy02)/(2g)

where h is the maximum height, vy0 is the initial vertical velocity, and g is the acceleration due to gravity. Since the ball is launched at an angle of 30.0 degrees above the horizontal, we can find vy0 using the equation:

vy0 = v0sin(θ)

where v0 is the initial velocity and θ is the launch angle. Finally, we can find v0 using the equation:

v0 = vy0/sin(θ)

To find the range, we can use the equation:

R = (v02sin(2θ))/g

where R is the range and θ is the launch angle.

Answer 2

a)

In order to find the initial velocity of the ball, we can use the formulas below:

`\begin{gathered} v_(y0)=v_0\cdot\sin (\theta) \\ v^2_y=v^2_(y0)+2\cdot g\cdot d \end{gathered}`

At the maximum height, the vertical speed is zero. So, using theta = 30°, d = 2.5 m and vy = 0, we have:

`\begin{gathered} v^2_y=v^2_(y0)+2\cdot g\cdot d \\ 0^2=v^2_{y0^{}}+2\cdot(-9.8)\cdot2.5 \\ v^2_(y0)=49^{} \\ v_(y0)=7 \\ \\ v_(y0)=v_y\cdot\sin (\theta) \\ 7=v_y\cdot(1)/(2) \\ v_y=14\text{ m/s} \end{gathered}`

b)

To find the range, let's calculate the horizontal component of the velocity and the time of flight:

`\begin{gathered} v_x=v\cdot\cos (\theta)_{} \\ v_x=14\cdot\frac{\sqrt[]{3}}{2} \\ v_x=12.124\text{ m/s} \\ \\ v_y=v_(y0)+g\cdot t \\ 0=7-9.8\cdot t \\ t=(7)/(9.8) \\ t=0.714 \\ \\ t_f=2t=1.43\text{ s} \\ \\ d_x=v_x\cdot t \\ d_x=12.124\cdot1.43 \\ d_x=17.34\text{ m} \end{gathered}`