Physics 1 Mechanics: Acceleration and Projectile Motion on a Ramp

[golfer1100]

A baseball is accelerated from rest at the rate a = 251 m/s/s along a rising ramp of lenth l=3m. The ramp makes an angle of theta =.5 rad with the horizontal. Neglect friction.

a)what is the launch velocity v knot of the ball? express v knot in a component form using a coordinate system of your choice. Sketch the ramp and show v knot and your coordinate system.

b) how far down the range x from the launch point does teh ball land?

c) suppose a wlal of heigh h stands at a distance d=x-31 m down range from the launch point. If the ball sails over the wall, waht is the maximum wall height?

 

[cronxeh]

Need to show your work, and also post in the correct forum

 

[golfer1100]

Im not sure where to start or where to post this

 

[cronxeh]

https://www.physicsforums.com/forumdisplay.php?f=153

How would you approach this problem? You are given acceleration in m/s^2 and length over which you accelerating, so you can find the speed at the end of the ramp now. Next you are given angles that gives you the distance that the ball will travel and stay airborne for. Just think about this problem, use appropriate formulas, and post if you get stuck

speed² = 2*acceleration*distance

 

[LudusRex]

a) The launch velocity of the ball can be calculated using the equation v = u + at, where u is the initial velocity, a is the acceleration, and t is the time. Since the ball starts from rest, u = 0. Therefore, v = at = (251 m/s/s)(3m) = 753 m/s. We can express this velocity in component form as v = (v_x, v_y), where v_x = v cosθ and v_y = v sinθ. Using the given angle of θ = 0.5 rad, we have v_x = (753 m/s) cos(0.5 rad) = 750.75 m/s and v_y = (753 m/s) sin(0.5 rad) = 6.25 m/s. The coordinate system can be chosen as x-axis along the ramp and y-axis perpendicular to it, as shown in the sketch below.

b) To calculate the range, we can use the equation x = ut + 1/2at^2, where x is the range, u is the initial velocity, a is the acceleration, and t is the time. Since the ball lands at the same height from which it was launched, we can use the time taken to reach the ground as the time of flight. The time can be calculated using the equation v = u + at, where u = 0 and v = 0 (since the ball reaches its maximum height and then falls back to the ground). Therefore, t = v/a = (0 m/s)/(251 m/s/s) = 0 seconds. Putting this value of t in the equation for range, we get x = (0 m/s)(0 seconds) + 1/2(251 m/s/s)(0 seconds)^2 = 0 m. So, the ball lands at a distance of 0 m from the launch point.

c) To calculate the maximum wall height, we can use the equation y = u_yt + 1/2at^2, where y is the height of the wall, u_y is the initial vertical velocity, a is the acceleration, and t is the time. We can calculate the time using the equation y = u_yt + 1/2at^2, where y = h, u_y = v_y = 6.25 m/s, and a = -9.8 m/s/s (since