Try and draw a free body diagram to help you understand this. The 35 m/s velocity is along the hypotenuse of a right triangle, and the components you want are in the x and y direction. To find the y component, you would use sinA=y/35, and for the x component, you would use cosA=x/35.
To reach the top of the arch, use the motion equations. You now have the velocity in the y direction, and you know acceleration due to gravity is 9.8 m/s2. Final velocity in the y direction at the top of the arch is 0, so use the equation vf2=vi2-2(9.8)(y) and solve for y.
The rock will be equal with your hand when the velocity in the y direction is equal to the initial y velocity, but in the opposite direction, so y=0. Use the equation y=vt-(1/2)(9.8)t2 to find the time it takes, then plug that time into the equation for movement in the x direction, which is x=v(in the x direction)t. Remember velocity in the x direction does not change in the absence of air resistance.
You attempt to throw the rock across the chasm with a velocity of 35 m/s at and angle of 60 degrees above horizontal. Calculate and label the forward velocity (components) of the rock as it leaves your hand. How long will it take to reach the top of the arc (apogee. How far away is the rock (from your hand) when is level with your hand the second time.
1 answer
0 answers