Suppose there is a cannon that shoots a cannon ball at a certain angle above the horizontal (a projectile). Since momentum is conserved in both directions, the cannon should posses a velocity now in the y (or z-axis if you would like to call it) and in the x axis.
1) However, what does it...
a) Work = Fd when they're in the same direction. So the force on the mass is mg (.105 x 9.8), and the distance it moves down is .06 meters. So solve for Fd (they're in the same direction so the answer is positive).
b) We'll write that the spring and earth do work on the mass: so -1/2kx^2 + mgh...
Break the motion into an x and y components. We'll say the swimmer undergoes a displacement of 480 meters in the +x direction. The time it takes for this to happen is 6 minutes and 40 seconds (400 seconds). So 480/400 = 1.2 m/s. Remember, the key to solving this is thinking about the problem...
So the soccer ball hits the crossbar (assuming it hits at 2.5 meters high). What does that mean? Do what fss said and break the motion into components, x and y. 2.5 is the positive y displacement. So we have 2.5 = vy (t) + 1/2a(t^2). So what is vy? and what is a? If you have a question...
Average Acceleration = Change in velocity / time. Let's first convert 1000km/hour to m/s. The answer to that is 277.7 meters/second. Now, we'll define direction (east is positive, west negative). So after it makes the turn, it's velocity is +277.7 m/s. Before it makes the turn, it's velocity...
In order to use the range formula, we would have to draw in the other half of the parabola so that the arrow ends up at the same height as it started (ie a vertical displacement of zero). This means we have the 66 m which is the half parabola it travels, plus the 66 m of the other half parabola...
In the original, you just did the math incorrectly: It's the F(up) - mg = ma. So F = ma + mg. In your coursebook, could you state the problem that it's listed for?
No, in my older post, i forgot to include the fact that the arrow was released parallel to the ground (ie no initial y velocity). I will re-write the corrected solution right here. If he releases the arrow parallel to the ground it makes half a parabola in its trajectory. Now, draw the other...
sorry, if he releases the arrow parallel to the ground it makes half a parabola in its trajectory. Now, draw the other half. You'll see that the x distance of this parabola now is 66 x2. Then, you can follow my last post. If you'd like to stick with this, you'd have to solve for the x...
Use the range equation for this, since it returns to the same initial height released (theoretically). V^2 Sin (2 theta)/g = Range. We know that the angle of 3 degrees it makes with the ground is the same three degrees it was released at above the horizontal. Hence, we have v^2 (sin 6)/9.8 =...
Since you have to assume that his y displacement is zero by the time he lands, you could use the range equation for this which is (v^2)sin(2x)/g. You know the range is 7.77 and you know x, which is the angle. So solve for v. For the second case, all is being kept constant, except for...
Thanks for your help; that's brilliant. However, wouldn't the stretch be the same even if it was lowered? For the first equation: x = mg/k. For the one slowly lowered, since the forces are balanced once it reaches equilibrium, shouldn't the stretch be x = mg/k also?