- #1

Parkkk41

- 2

- 0

1. If an object is thrown with a horizontal velocity of 32 m/s and a vertical velocity of 24 m/s from the top of a cliff that is 230 m in height (on the moon where the acceleration is found to be 1.6 m/s^2), how long will the object be in flight?

2. Δd = v1Δt+1/2aΔt^2, Δd = v2Δt-1/2aΔt^2, x = (-b±√(b^2-4ac))/2a

3. Given information: a = -1.6 m/s^2, Δd = 320m, Vv = 24 m/s, Δt = ?

Δd = v1Δt+1/2aΔt^2

320 = 24Δt+0.5(-1.6)(Δt^2)

320 = 24Δt+0.8Δt^2

0 = 0.8Δt^2+24Δt-320

----> Solve for zeros using quadratic formula

x = (-b±√(b^2-4ac))/2a

= -24±√(24^2-4(0.8)(-320))/2(0.8)

=(-24+40)/1.6 = (-24-40)/1.6

=10s =-40s

So, -40 would not be used and 10 is used... But the thing is, my teacher says the answer is 20? Meaning the half flight is 20s so multiply it by 2 and it's in flight for a total of 40s... I do not understand how 40s is the answer. Can someone please tell me how to do this right or verify the real answer? Any help is appreciated, thank you :)

2. Δd = v1Δt+1/2aΔt^2, Δd = v2Δt-1/2aΔt^2, x = (-b±√(b^2-4ac))/2a

3. Given information: a = -1.6 m/s^2, Δd = 320m, Vv = 24 m/s, Δt = ?

Δd = v1Δt+1/2aΔt^2

320 = 24Δt+0.5(-1.6)(Δt^2)

320 = 24Δt+0.8Δt^2

0 = 0.8Δt^2+24Δt-320

----> Solve for zeros using quadratic formula

x = (-b±√(b^2-4ac))/2a

= -24±√(24^2-4(0.8)(-320))/2(0.8)

=(-24+40)/1.6 = (-24-40)/1.6

=10s =-40s

So, -40 would not be used and 10 is used... But the thing is, my teacher says the answer is 20? Meaning the half flight is 20s so multiply it by 2 and it's in flight for a total of 40s... I do not understand how 40s is the answer. Can someone please tell me how to do this right or verify the real answer? Any help is appreciated, thank you :)

Last edited by a moderator: