Respuesta :
Answer:
The magnitude of the launch velocity is 149.3 m/s
The distance is 74.65 m
Explanation:
Given that,
Angle = 24°
Time =12.5 sec
We need to calculate the magnitude of the launch velocity
Using equation of motion
[tex]y=-\dfrac{1}{2}gt^2+ut[/tex]
[tex]y=-\dfrac{1}{2}gt^2+u\sin\theta\times t[/tex]
Put the value in to the formula
[tex]0=-\dfrac{1}{2}\times9.8\times(12.5)^2+u\sin24\times12.5[/tex]
[tex]u=\dfrac{765.625}{0.41\times12.5}[/tex]
[tex]u=149.3\ m/s[/tex]
We need to calculate the distance
Using formula of distance
[tex]x=u\cos\theta\times t[/tex]
Put the value into the formula
[tex]x=149.3\cos24\times12.5[/tex]
[tex]x=74.65\ m[/tex]
Hence, The magnitude of the launch velocity is 149.3 m/s
The distance is 74.65 m
