Answer:
Explanation:
<u>Given:</u>
velocity of the ball first and ball second, 
angle of projection of the first ball, 
∵The balls should land at the same point,
∴their range of projectile,
As we know for the range of projectile:
∵ we have equal range in both the cases
![2\theta_2=sin^{-1}[sin 144^{\circ}]](https://tex.z-dn.net/?f=2%5Ctheta_2%3Dsin%5E%7B-1%7D%5Bsin%20144%5E%7B%5Ccirc%7D%5D)
![\theta_2=\frac{1}{2}\times sin^{-1}[sin 144^{\circ}]](https://tex.z-dn.net/?f=%5Ctheta_2%3D%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20sin%5E%7B-1%7D%5Bsin%20144%5E%7B%5Ccirc%7D%5D)
Answer:
V = 0.248 L
Explanation:
To do this, use the following equation:
P1*V1/T1 = P2*V2/T2
This equation is used to find a relation between two differents conditions of a same gas, which is this case. From this equation we can solve for V2.
Solving for V2:
V2 = P1*V1*T2/T1*P2
Temperature must be at Kelvin, so, we have to sum the temperature 273 to convert it in K.
Replacing the data we have:
V2 = 1 * 4.91 * (-196+273) / 5.2 * (20+273)
V2 = 378.07 / 1523.6
V2 = 0.248 L
