If a projectile is fired straight up at a speed of 10 m/s, the total time to return to its starting position is about ?
-Hfinal=hinitial+ vinitial*t -4.9t^2
0=0+10t-4.9t^2 when I solved for t I got .49t. Also, where did 4.9 come from?
A man leans over the edge of a cliff and throws a rock upward at 4.9m/s. How far below the level from which it was thrown is the rock 2 seconds later?
-Hfinal=Hinitial+Vi*t - 4.9t2
hfinal=0+4.9m/s*2 s -4.9t^2, then I get hfianl=9.8m/s/s-4.9^2, I don't dont what to do next?
4.9 is g/2, and g is the acceleration of gravity on Earth. The answer should be in seconds, not 0.49 t. You also did the calculation incorrectly.
In your second problem, the equation to solve is
y = 4.9t - 4.9 t^2. Plug in t = 2s and solve for y.
y(at t=2s) = 9.8 - 4*9.8 = -29.4 m
To find the total time it takes for a projectile to return to its starting position, you can use the equation of motion:
Hfinal = Hinitial + Vinitial * t - 4.9 * t^2
In this equation, Hfinal represents the final height, Hinitial represents the initial height, Vinitial represents the initial velocity, t represents time, and 4.9 is the acceleration due to gravity, which is approximately 9.8 m/s^2, divided by 2 because the projectile is moving in both the upward and downward direction.
For the first question, the projectile is fired straight up with an initial velocity of 10 m/s. Since it returns to its starting position, the final height (Hfinal) is the same as the initial height (Hinitial), which is 0. Plugging these values into the equation:
0 = 0 + 10t - 4.9t^2
To solve for t, rearrange the equation into a quadratic form and solve using the quadratic formula or factoring. In this case, rearranging the equation gives:
4.9t^2 - 10t = 0
Simplifying the equation further gives:
t(4.9t - 10)= 0
This equation has two solutions: t = 0 (which is the initial time when the projectile is fired) and (4.9t - 10) = 0. Solving the second part of the equation gives:
4.9t - 10 = 0
4.9t = 10
t = 10 / 4.9
t ≈ 2.04 seconds
Therefore, it takes about 2.04 seconds for the projectile to return to its starting position.
In the second question, the rock is thrown upward with an initial velocity of 4.9 m/s. We want to find how far below the level it was thrown after 2 seconds.
Using the same equation:
Hfinal = Hinitial + Vinitial * t - 4.9 * t^2
Substituting the given values:
Hfinal = 0 + (4.9 m/s) * (2 s) - 4.9 * (2 s)^2
Simplifying:
Hfinal = 0 + 9.8 m/s - 4.9 * 4 s^2
Hfinal = 0 + 9.8 m/s - 19.6 s^2
To evaluate this equation further, we need the value of s^2. However, we don't have enough information to calculate it in this scenario. It seems there might be an error in your calculations.
To find the total time for the projectile to return to its starting position, we can use the equation:
Hfinal = Hinitial + Vinitial*t - 4.9t^2
where Hfinal is the final height, Hinitial is the initial height (which is usually 0 in this case since the projectile starts from the ground), Vinitial is the initial velocity (which is 10 m/s in this case), t is the time.
Substituting the values into the equation:
0 = 0 + 10t - 4.9t^2
To solve for t, we can rearrange the equation:
4.9t^2 - 10t = 0
Factoring out a t from both terms:
t(4.9t - 10) = 0
Setting each factor equal to zero:
t = 0 (discard this since it represents the initial time at t = 0)
4.9t - 10 = 0
Solving for t:
4.9t = 10
t = 10 / 4.9
Calculating the value:
t ≈ 2.04 seconds
So, the total time for the projectile to return to its starting position is approximately 2.04 seconds.
Now, for the second question:
To find how far below the level from which the rock was thrown 2 seconds later, we can again use the equation:
Hfinal = Hinitial + Vinitial*t - 4.9t^2
Given that Hinitial is the initial height (which is the height from the cliff edge in this case), Vinitial is the initial velocity (which is 4.9 m/s upward in this case), and t is the time (which is 2 seconds in this case).
Substituting the values into the equation:
Hfinal = 0 + (4.9 m/s * 2 s) - 4.9(2^2)
Simplifying:
Hfinal = 9.8 m/s - 19.6 m
Hfinal = -9.8 m
The negative sign indicates that the rock is below the starting level, and the magnitude represents the distance below. Therefore, the rock is 9.8 meters below the level from which it was thrown 2 seconds later.