This problem is going to be pretty long to solve. So, prepare.
We’re interested in the change in our x position. So we have to break the velocity vector up into its components. Do cosine of 50 and then multiply by the magnitude of the velocity. I got 20.57m/s. That’s our initial velocity. And remember, horizontal acceleration is zero. The vertical acceleration, or any vertical component, has no effect on the horizontal components. In order to solve this problem, we want to utilize this equation:
Change in x-position = Vix*t
Let’s solve for time, which is dependent on the vertical components. The projectile will stop when it vertically hits the ground. Generally you want to use this equation for solving for time:
Yf = Yi + Viy*t + 1/2at^2
We didn’t solve for the vertical component yet, so let’s do that now. (Sine of 50)*(32) = 24.51m/s
Let’s now plug everything in:
0 = 0 + 24.5t - 4.9t^2
0 = 24.5t - 4.9t^2
0 = t(24.5 - 4.9t)
-24.5 = -4.9t
t = 5 seconds
The hard stuff is pretty much over. Put that 5 seconds into the other equation I said we wanted to use to solve the problem
Change in x-position (range) = (20.57)*(5)
= 102.85 meters
Answer B
Force can be expressed as the product of mass and acceleration. Mathematically, that's F = m(a). Plugging the given into the equation, we have F = (13.5 kg)(9.5 m/s²) = 128.3 kg.m/s² or 128.3 N<span>. </span>
Answer:
Energia cinetica=187.5 [J]
Explanation:
La energia cinetica se define por medio de la siguiente ecuacion:
Debemos convertir las unidades de kilómetros por hora a metros por segundo
Reemplazamos en la ecuacion de energia cinetica.
Answer:
4.5 m/s
Explanation:
The rock must barely clear the shelf below, this means that the horizontal distance covered must be
while the vertical distance covered must be
The rock is thrown horizontally with velocity , so we can rewrite the horizontal distance as
where t is the time of flight. Re-arranging the equation,
(1)
The vertical distance covered instead is
where we omit the term since the initial vertical velocity is zero. From this equation,
(2)
Equating (1) and (2), we can solve the equation to find :