Mechanics; grade 12 Projectile motion

[ttt6683]

Summary:: Scooby-Doo wants a Scooby Snack. According to a website I looked up last night, Scooby can run at a speed of 36 km/h. He runs up a ramp (as shown, 2.0 meters high, 4.0 meters long) at that speed and off the edge. Three meters from the end of the ramp, Shaggy has a Scooby-Snack launcher which will fire a snack straight up. At the instant Scooby leaves the ramp, Shaggy fires the snack. What is the initial speed of the launched snack for Scooby to catch it in midair? [6]

Not sure if this correct. How do you find the X and Y components with no given angle, so I tried to find the angle of the hill.

[Moderator's note: Moved from a technical forum and thus no template.]

 

[PeroK]

How about ignoring gravity, as it is the same for Scooby and for the snack?

And, instead of calculating $\theta$, why not take $u_x = 2u_y$?

 

[Lnewqban]

Welcome, ttt6683!

Your calculated angle for the ramp seems to be incorrect.
Therefore, your calculated values of $V_x$ and of $V_y$ for Scooby-Doo at the edge of the ramp are not correct either.

 

[PeroK]

Welcome, ttt6683!

Your calculated angle for the ramp seems to be incorrect.
Therefore, your calculated values of $V_x$ and of $V_y$ for Scooby-Doo at the edge of the ramp are not correct either.

Which is ironic, given there was no need to calcuate the angle!

Moral: avoid unnecessary calculations.

 

[PeroK]

Not sure if this correct. How do you find the X and Y components with no given angle, so I tried to find the angle of the hill.

A neat way to do this is to observe that $$\frac {u_y}{u_x} = \tan \theta = \frac 2 4$$, hence $$u_x = 2u_y$$
And:
$$u^2 = u_x^2 + u_y^2 = 5u_y^2$$ hence $$u_y = \frac u {\sqrt{5}} \ \ \text{and} \ \ u_x = \frac{2u}{\sqrt{5}} $$
Where $u = 10m/s$ is the initial launch speed.