Answer:
[tex]v_{0x} = 10.90\ m/s[/tex]
Explanation:
given,
time = 0.7 s
displacement of the car = r = 8 m
vertical displacement of the car
using equation of motion
[tex]y= ut + \dfrac{1}{2}gt^2[/tex]
initial velocity of the car is equal to zero
[tex]y=\dfrac{1}{2}gt^2[/tex]
now, horizontal distance traveled by the car
r² = x² + y²
[tex]x = \sqrt{r^2-y^2}[/tex]
[tex]v_{0x} =\dfrac{x}{t}[/tex]
[tex]v_{0x} =\dfrac{\sqrt{r^2-y^2}}{t}[/tex]
[tex]v_{0x} =\dfrac{\sqrt{r^2-(\dfrac{1}{2}gt^2)^2}}{t}[/tex]
[tex]v_{0x} =\dfrac{\sqrt{8^2-(\dfrac{1}{2}(-9.8)\times 0.7^2)^2}}{0.7}[/tex]
[tex]v_{0x} = 10.90\ m/s[/tex]
