Complete Question
The diagram illustrating this question is shown on the first uploaded image
Answer:
The value is [tex]\theta = 27^o[/tex]
Explanation:
Generally the distance covered by the ball thrown at angle 45° that did not bounce is mathematically represented as
[tex]D = v cos (45) * t[/tex]
Here t is the total time duration which is mathematically represented as
[tex]t = \frac{v sin (45)}{g}[/tex]
This distance distance is also mathematically represented as
[tex]D = vcos(\theta)* t_1 + \frac{v}{2} cos(\theta )*t_2[/tex]
Here [tex]\theta[/tex] the angles made as shown in the diagram
Here [tex]t_1[/tex] is the time before the first bounce which is mathematically represented as
[tex]t_1 = \frac{v sin (\theta )}{g}[/tex]
[tex]t_1[/tex] is the time duration before the final point which is mathematically represented as
[tex]t_2 = \frac{\frac{v}{2} * sin (\theta )}{g}[/tex]
So
[tex]v cos (45) * \frac{v sin (45)}{g} = vcos(\theta)* \frac{v sin (\theta )}{g} + \frac{v}{2} cos(\theta ) * \frac{\frac{v}{2} * sin (\theta )}{g}[/tex]
=> [tex]cos (45) sin(45) = cos(\theta)sin(\theta ) + cos(\theta )sin(\frac{\theta}{4})[/tex]
=> [tex]0.5 = \frac{5}{8} * sin(2\theta )[/tex]
=> [tex]\theta = 27^o[/tex]
