# Another projectile questions involves distance

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• May 26th 2011, 09:27 AM
silvercats
Another projectile questions involves distance
Object is projected from a speed of V and an angle of Θ
Horizontal displacement is 15m ,vertical displacement is 5m
Find V and Tan Θ

(right )
x=V cos Θ*t
t=x/V cos Θ

(up)
y=v Sin Θ*t-g/2*t²
from first equation y=V sinΘ(x/V cosΘ)-g/2*(x/v cosΘ)²
5=V sin Θ(15/V cos Θ)-g/2 *(15/V cosΘ)² <<<Stuck!!!!
help wanted

Books says 12.5 , 4/3
• May 26th 2011, 10:20 AM
abhishekkgp
Quote:

Originally Posted by silvercats
Object is projected from a speed of V and an angle of Θ
Horizontal displacement is 15m ,vertical displacement is 5m
Find V and Tan Θ

(right )
x=V cos Θ*t
t=x/V cos Θ

(up)
y=v Sin Θ*t-g/2*t²
instead of using 't' you should have used 't/2' in the above eqn. do you see why?
from first equation y=V sinΘ(x/V cosΘ)-g/2*(x/v cosΘ)²
5=V sin Θ(15/V cos Θ)-g/2 *(15/V cosΘ)² <<<Stuck!!!!
help wanted

Books says 12.5 , 4/3

did this help?
• May 26th 2011, 10:32 AM
silvercats
Quote:

Originally Posted by abhishekkgp
instead of using 't' you should have used 't/2' in the above eqn. do you see why?

why?????
• May 26th 2011, 12:11 PM
abhishekkgp
Quote:

Originally Posted by silvercats
why?????

't' is the total time of flight, that is the time taken by the projectile to hit the ground. it takes half the time of flight to achieve maximum height(here max height is 5m). now do you understand what i was trying to say?
• May 26th 2011, 07:27 PM
silvercats
ah.. yeah.can you guide me through the rest?
thanks