At time t = 0 , a ball is struck at ground level and sent over level ground. The momentum p versus t during the flight is given by Figure (with p₀ = 6.0 kg.m/s and p₁ = 4.0 kg.m/s). At what initial angle is the ball launched? (Hint: find a solution that does not require you to read the time of the low point of the plot.)

The initial momentum of the ball is$p_0=6.0kh.m/s$

The final momentum of the ball is$p_1=4.0kh.m/s$

Hint: Find the solution that does not require you to read the time of the low point of the plot.

Trigonometry, a branch of mathematics dealing with specific functions of angles and their application in calculations. There are six angle functions commonly used in trigonometry. Their names and abbreviations are sine (sin), cosine (cos), tangent (tan), cotangent (cot), secant (sec), and cosecant (csc).

You can use the concept of trigonometry.

Formula:

$\cos \theta =\frac{Adjacentside}{Hypotenuse}$

From the graph given, you can conclude that horizontal component of the momentum is,

$p_1=4.0kg.m/s$

This component would be constant throughout the motion. This is because velocity in the horizontal direction would not be affected by gravity. So, at t = 0 , you have the horizontal component of the momentum as,

$p_1=4.0kg.m/s$

And total momentum as,

$p_0=6.0kg.m/s$

Therefore, using trigonometry, at t = 0 you can write,

$$\begin{gathered} \cos \theta =\frac{p_1}{p_0} \\ =\frac{4.0kg.m/s}{6.0kg.m/s} \\ \theta ={\cos }^{-1}\left(0.67\right) \\ =48° \end{gathered}$$

Hence, the initial launching angle of the ball by using trigonometry is $48°$.