Simple projectile spring-loaded plunger motion

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A spring-loaded plunger launches a ball at 3.50 m/s from a 20-degree inclined board, aiming to hit a target 2.45 m away. To solve the problem, it's essential to analyze the time it takes for the ball to travel horizontally and vertically, using kinematic equations for both motions. The key is to set the time equations for vertical and horizontal motion equal to each other to find the angle theta. A trigonometric identity involving sin(2*theta) may be useful in determining the correct angle. Understanding the independence of horizontal and vertical motions is crucial for solving this projectile motion problem effectively.
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i don't know why I am having trouble on this...

In one contest at the county fair, a spring-loaded plunger launches a ball at a speed of 3.50 m/s from one corner of a smooth, flat board that is tilted up at a 20 degree angle. To win, you must make the ball hit a small target at the adjacent corner, 2.45 m away. At what angle theta should you tilt the ball launcher?

knight_Figure_06_33.jpg


a along the y-axis is: 3.3518m/s^2

i don't know what to do next... please help

i know that youre supposed to look at it when it reaches the top of its arc when

vf=0m/s
vi=3.5m/s
d= 1.225m
t=?
a(ball)=?
a (y-axis): 3.3518m/s^2
theta=?
 
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Often with this kind of problem, it helps to think about the time it's going to take for the object to get somewhere. How long will it take for the ball to make it across the board (depends on the angle)? You need to make the ball go across the table in the same amount of time it takes it to go up the incline and all the way back down. You'll end up with two simultaneous equations, and just solve for the time. That will take you back to the angle solution...
 
I have the same problem as this except the ball is launched at 3.0m/s and the target is 2.12m away.
What do I need to find out to solve this? I'm trying to come up with the formulas to find the answer but I just keep hitting dead ends.
 
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The only kinematic equations you need are:

x_f = x_i +v_ix*t+0.5*a_x*t^2
y_f = y_i +v_iy*t+0.5*a_y*t^2

Keep in mind that horizontal motion and vertical motion are completely independent of one another (one of the convenient things about projectile motion).

With your origin set as the bottom left-hand corner of the ramp--

Find t in terms of theta for when y equals 0 (the 2nd time: at the other end of the parabola), and then find t in terms of theta for when your x position is at its desired maximum. Set one t equal to the other and solve - that will give you the intersection for the 2 t equations you came up with.

When I worked through the problem, I had to use a trigonometric identity to come up with the correct value for theta. (Hint: sin(2*theta).)
 
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