Answer:
The distance from the building is 11.41 m.
Explanation:
Given that,
Speed = 30.0 m/s
Angle = 45.0°
Distance = 10.0 m
We need to calculate the vertical component
Using formula of vertical component
[tex]u_{y}=u\sin\theta[/tex]
[tex]u_{y}=30.0\times\sin45[/tex]
[tex]u_{y}=21.21\ m/s[/tex]
We need to calculate the time
Using equation of motion
[tex]s=ut-\dfrac{1}{2}gt^2[/tex]
Put the value into the formula
[tex]10=21.21t-\dfrac{1}{2}\times9.8\times t^2[/tex]
[tex]21.21t-\dfrac{1}{2}\times9.8\times t^2-10=0[/tex]
[tex]-4.9t^2+21.21t-10=0[/tex]
[tex]t=0.538\ sec[/tex]
On neglecting of higher value of t
Then use the horizontal component
[tex]u_{y}=u\cos\theta[/tex]
[tex]u_{x}=30.0\times\cos45[/tex]
[tex]u_{x}=21.21\ m/s[/tex]
We need to calculate the distance
[tex]d=v\times t[/tex]
Put the value into the formula
[tex]d=21.21\times0.538[/tex]
[tex]d=11.41\ m[/tex]
Hence, The distance from the building is 11.41 m.
