Find time given initial velocity, distance and acceleration.

[physicN00Bz]

Homework Statement

Coyote falling down with a parachute holding an anvil at 10m/s. At 50m he drops the anvil.

A) How long does it take for the anvil to hit the ground?
B) what is the velocity of the anvil right before it hits the ground?

Homework Equations

X= Xo + V0T+1/2A(T^2)

g= 9.8 m/s^2

The Attempt at a Solution

50m = 10m/sT + 4.9m/s^2T^2

i use the quadratic equation and got t= 1.14 and -2.143

however i know this is wrong since i can not get a negative second and it seen like the "s" cancel out and you got T = a number without second.

also i know the answer is 2.33 s for (A) but want to know how to get that answer.

 

[Bhumble]

Seems like you've done everything correctly except solving the quadratic. I get 2.333 and -4.37381 for my solutions. Or at least my calculator does...

 

[physicN00Bz]

Seems like you've done everything correctly except solving the quadratic. I get 2.333 and -4.37381 for my solutions. Or at least my calculator does...

yeah i actually re did the quadratic i also got 2.33 and -4.373

 

[physicN00Bz]

ok for

B) what is the velocity of the anvil right before it hits the ground?

i use V = Vo + AT and got V= 10m/s + 9.8 m/s^2(2.33s) got V =32.9 m/s

However the answer is in negative (as in -32.9 m/s) what did i do wrong to miss the negative sign?

 

[rwisz]

To solve for the time it takes for the anvil to hit the ground, we can use the formula X = Xo + V0T + 1/2AT^2, where X is the final position (50m), Xo is the initial position (0m), V0 is the initial velocity (10m/s), and A is the acceleration (9.8m/s^2).

Substituting in the given values, we get:

50m = 0m + (10m/s)T + 1/2(9.8m/s^2)T^2

Rearranging and simplifying, we get:

4.9T^2 + 10T - 50 = 0

Using the quadratic formula, we get two possible solutions for T:

T = (-10 ± √(10^2 - 4(4.9)(-50)))/(2(4.9))

T = (-10 ± √(100 + 980))/9.8

T = (-10 ± √(1080))/9.8

T ≈ (-10 ± 32.91)/9.8

T ≈ -3.3 or 2.33

Since time cannot be negative, we can discard the negative solution and conclude that it takes approximately 2.33 seconds for the anvil to hit the ground.

To find the velocity of the anvil right before it hits the ground, we can use the formula V = V0 + AT, where V is the final velocity, V0 is the initial velocity, A is the acceleration, and T is the time.

Substituting in the known values, we get:

V = 10m/s + (9.8m/s^2)(2.33s)

V ≈ 10m/s + 22.83m/s

V ≈ 32.83m/s

Therefore, the velocity of the anvil right before it hits the ground is approximately 32.83m/s.