Respuesta :
Just consider the vertical component
[tex]d_y=v_{0,y}t+\frac{1}{2}at^2[/tex]
[tex]d_y=(13.9\sin 25\°)t+\frac{1}{2}(-9.8~m/s^2)t^2 \\ \\ d_y=5.87t-4.9t^2[/tex]
Set d_y equal to 0 to find the times the ball is on the ground
[tex]0=5.87t-4.9t^2 \\ \\ 0=t(5.87-4.9t) \\ \\ t=0 \\ \\ t=1.2~s[/tex]
The ball was in the air for 1.2 seconds
[tex]d_y=v_{0,y}t+\frac{1}{2}at^2[/tex]
[tex]d_y=(13.9\sin 25\°)t+\frac{1}{2}(-9.8~m/s^2)t^2 \\ \\ d_y=5.87t-4.9t^2[/tex]
Set d_y equal to 0 to find the times the ball is on the ground
[tex]0=5.87t-4.9t^2 \\ \\ 0=t(5.87-4.9t) \\ \\ t=0 \\ \\ t=1.2~s[/tex]
The ball was in the air for 1.2 seconds
Answer:
The 1st one is 1.2 and the Second one is 15.1
Explanation:
Just did it on Quiz and got it correct.