A pumpkin is thrown upwards at a speed of 27.5 m/s and from a height of 2 m above the ground. How long before it hits the ground and smashes?
Not sure where to begin with this.
h = Hi + Vi t - 4.9 t^2
0 = 2 + 27.5 t -4.9 t^2
or
4.9 t^2 -27.5 t - 2 = 0
solve quadratic for t, use the positive answer (any negative root was on the way up before you threw it :)
V^2 = Vo^2 + 2g*h.
0 = 27.5^2 - 19.8h
h = 38.6m. above launching point.
ho + h = 2 + 38.6 = 40.6m above gnd.
V = Vo + g*Tr.
0 = 27.5 - 9.8Tr
Tr = 2.81s. = Rise time.
0.5g*Tf^2 = 40.6.
Tf(fall time) = ?.
Tr+Tf = Time to reach gnd.
To solve this problem, you can use the kinematic equation for vertical motion:
h = h0 + v0*t - (1/2)gt^2
Where:
- h is the height of the pumpkin above the ground at any given time.
- h0 is the initial height (2 m in this case).
- v0 is the initial velocity (27.5 m/s in this case).
- g is the acceleration due to gravity (approximately 9.8 m/s^2).
- t is the time.
We want to find the time it takes for the pumpkin to hit the ground, so we need to solve for t when h equals 0 (because the pumpkin hits the ground when its height is zero).
0 = 2 + 27.5t - (1/2)(9.8)t^2
To solve this quadratic equation, we need to rearrange it and set it equal to zero:
-4.9t^2 + 27.5t + 2 = 0
Now we can solve this equation using the quadratic formula:
t = (-b ± √(b² - 4ac)) / (2a)
In our case, a = -4.9, b = 27.5, and c = 2.
t = (-27.5 ± √(27.5² - 4(-4.9)(2))) / (2 * -4.9)
Simplifying this expression will give us the two solutions, but we know that time cannot be negative in this problem since we are only interested in the time before the pumpkin hits the ground. So we only consider the positive value of t.
t = (-27.5 + √(27.5² - 4(-4.9)(2))) / (2 * -4.9)
By evaluating this expression, you will find the positive value of t, which represents the time it takes for the pumpkin to hit the ground and smash.