Projectile motion doubt (1 Viewer)

[prazo94]

Hi guys, could you help me with this question:
A ball is thrown on level ground at an initial speed of V m/s at an angle of projection of a(alpha)
Assume that t seconds after release, the horizontal and vertical displacement are given by
x= Vtcosa while y=VtSina- 1/2gt^2

I've shown that the equation of the horizontal range is V^2 sin2a/g as well

So the question is: when V=30m/s, the ball lands 45 metres away. Take g=10m/s
i) find two possible values of a(alpha): i've found this to be 15 and 75

ii) A 2 metre high fence is placed 40metres from the thrower. Examine each trajectory to see whether the ball will still travel 45metres?(how would do you
prove this? what equations do we use?)

thanks!

 

[Timske]

I got a = 15 and 75 as well

 

[prazo94]

^could you do part ii) of the question?

 

[Sy123]

Ok, so what are the conditions for it to go over the fence and still reach 45m range?
It is, when x=40 y must be greater than or equal to 2. While a is 15 or 75 degrees.

So lets find the time when x=40 (keeping in mind V=30 and g=10)



Lets sub this time into our y-eqn in order to find the y-value when x=40 and then sub in our alpha such that range is still 45