Question in dynamics projectile motion

[abdo99]

Attempt solution

Y=(vsinø)t-0.5gt^2
4=(vsinø)t-4,9t^2

But there is 3 variables!

Another eq is
X=(vcosø)t
6=(vcosø)t

But i did not found any away to solution

 

[dauto]

You posted a sideways picture

 

[rtsswmdktbmhw]

Hint: 1. The ball travels on a parabolic path.

2. If you know the max height of the ball, you can calculate the vertical and horizontal components of initial velocity.

Or, at least, that's how I managed to do it but there may be a simpler method/formula..

 

[abdo99]

I know that ,but i did not konw the maximum H
The rule is H=(u^2sin^2)/2g

 

[HalfLostAndSearching]

In order to solve this problem, we need to use the kinematic equations for projectile motion. These equations relate the displacement (x and y), initial velocity (v), initial angle (ø), time (t), and acceleration due to gravity (g). In this case, we have two equations for y and x, but we need to find the values for v, ø, and t.

To do this, we can use a system of equations, where we set the two equations equal to each other and solve for v and t. This will give us the initial velocity and time at which the projectile reaches a height of 4 meters and a horizontal distance of 6 meters.

Once we have the values for v and t, we can then use those values to solve for ø using the equation ø = tan^-1 (y/x). This will give us the initial angle at which the projectile was launched.

However, it's important to note that in order to fully solve this problem, we also need to know the initial position of the projectile and the angle at which it was launched. Without this information, we cannot fully determine the solution.