1. A stone is tossed upward with a velocity of 8m/s from the edge of a cliff 63 m high. How long will it take the stone to hit the ground at the foot of the cliff?
Using antidifferentiation method, I got dv=-9.8dt. The I integrated and got v+constant1=-9.8t+constant2. Then simplified to v=-9.8t+constant. Then I plugged 9m/s into v, and 63 into constant to solve for t, which is 5.6s. My answer is wrong, and I don't know how to solve this.
a(t) = -9.8
v(t) = -9.8t + c
Initial Velocity = -8m/s. So,
-8 = -9.8(0) + c
-8 = c
So, v(t) = -9.8t - 8
s(t) = -4.9t^2 - 8t + d
Initial Position = 63m
63 = -4.9(0)^2 - 8(0) + d
63 = d
So, s(t) = -4.9t^2 - 8t +63
0 = -4.9t^2 - 8t +63
t approximately equals = 2.86 or 2.9 seconds
(verified with the textbook answer)
To solve this problem, you can use the equations of motion. Let's go through the steps together:
1. Identify the known values:
- Initial velocity (u): 8 m/s (upward)
- Acceleration due to gravity (g): -9.8 m/s^2 (negative sign indicates downward direction)
- Displacement (s): -63 m (negative sign indicates downward direction)
2. Choose an appropriate equation of motion:
- In this case, we can use the equation: s = ut + (1/2)gt^2
3. Plug in the known values into the equation:
- -63 = (8)t + (1/2)(-9.8)t^2
4. Rearrange the equation:
- Rewrite the equation in standard quadratic form: (1/2)(-9.8)t^2 + (8)t - 63 = 0
5. Solve the quadratic equation:
- You can solve this equation using the quadratic formula: t = (-b ± sqrt(b^2 - 4ac))/(2a)
- Substitute a = (1/2)(-9.8), b = 8, and c = -63 into the formula.
6. Calculate the time(s):
- Solve for t using the quadratic formula and find the positive value.
By following these steps, you will obtain the correct answer for the time it takes for the stone to hit the ground at the foot of the cliff.
In addition, could I also get help with this substitution question?
(y^3+6y^2+12y+8)(y^2+4y+4)dy
I have no idea. I made the first expression into u, derived it, plugged it back in to get 1/3|integration|(u)du. I don't know how to go on from here.
we know a = -9.8
v = -9.8t + c
when t = 0 , v = 8
8 = 0+c --> c = 8
v = -9/8t + 8
s = -4.9t^2 + 8t + 63 , where s is the distance in metres.
at hitting the ground, s = 0
-4.9t^2 + 8t + 63 = 0
t = (-8 ± √1298.8)/-9.8
= appr 4.49 seconds or a negative t