an orangutan throws a coconut vertically upward at the foot of a clif 40 m high while his mate simulaneously drops another coconut from the top of the clif. the two coconuts collide at an altitude of 20 m what is the initial velocity of the coconut that was thrown upward?
i know the equations
x= position = .5a t^2 + v (velocity) initial t + x initial (pos)
v= at+v initial
a=a
please help thanks!!
Go to the bottom coconut
20=Vi*t-1/2 g t^2
Now go to the top coconut
20=40-1/2 g t^2
solve for t in the second equation, put it in the first, solve for vi
waht is g?
and what do you mean by top and bottom?
thanks
top of clift, bottom of cliff
g is -9.8m/s^2, the acceleration of gravity.
why didyou use
these equations
Go to the bottom coconut
20=Vi*t-1/2 g t^2
Now go to the top coconut
20=40-1/2 g t^2
You were given distance, those are the distance equation.
finalposition=initial position+vi*t+ 1/2 a*t^2
memorize that. g= is negative(downward), so make the last term - 1/2 9.8 t^2, some just write - 4.9t^2
If objects acted simultaneously their relationship is time. The time is obtain because the time of collision between the coconut is the same.
A coconut is hanging on a tree at height of 15m from the ground a body launches as projectile vertically upward with velocity 20m/sat what time projectile will pass the coconut
To find the initial velocity of the coconut thrown upward, we can start by setting up equations for both coconuts.
For the coconut thrown upward:
We know the final altitude of the coconut when it collides is 20 m. Let's assign this as its final position, x final.
The initial altitude of the coconut is 0 m, which we can assign as x initial.
The acceleration due to gravity, a, is -9.8 m/s^2 because the coconut is moving against gravity.
We are looking for the initial velocity of the coconut thrown upward, which we can assign as v initial.
Using the equation for position, x = 0.5 a t^2 + v initial t + x initial,
we can plug in the known values to get: 20 = 0.5 * (-9.8) * t^2 + v initial * t + 0.
Similarly, for the coconut dropped from the top of the cliff:
The final altitude of the coconut is also 20 m, which is its initial position, x initial.
The initial altitude of the coconut is the total height of the cliff, which is 40 m, assigned as x initial.
Again, the acceleration due to gravity, a, is -9.8 m/s^2.
The initial velocity of the coconut is 0 m/s because it was dropped from rest.
Using the same equation: 20 = 0.5 * (-9.8) * t^2 + 0 * t + 40,
Now we have two equations with two unknowns:
Equation 1: 20 = -4.9 * t^2 + v initial * t,
Equation 2: 20 = -4.9 * t^2 + 40.
Subtracting Equation 2 from Equation 1, we can eliminate the t^2 term:
0 = v initial * t - 40.
Since the coconuts collide mid-air, they both take the same time to reach the point of collision. Thus, the value of t is the same for both equations.
Now, we can solve for t by rearranging the equation:
v initial * t = 40,
t = 40 / v initial.
Substituting this value of t into the first equation:
0 = (40 / v initial) * v initial - 40,
0 = 40 - 40,
0 = 0.
This equation always holds true because the left side is zero. Therefore, the value of v initial does not matter.
Hence, the initial velocity of the coconut thrown upward can be any value, and the coconuts will still collide at an altitude of 20 m.