# Projectile Motion Problem (Object thrown from the top of a cliff)

• Parkkk41
If the ball is thrown horizontally, the answer is 20 seconds.In summary, the time in flight can be determined using the equation Δd = v1Δt+1/2aΔt^2, where Δd is the displacement, v1 is the initial velocity, a is the acceleration, and Δt is the time. By substituting the given values, the equation becomes 320 = 24Δt+0.5(-1.6)(Δt^2). This can be solved using the quadratic formula, giving two solutions of 10 seconds and -40 seconds. However, since the object cannot be in flight for a negative amount of time, the solution of -40 seconds

#### Parkkk41

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 :)

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Parkkk41 said:
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 :)
Hello Parkkk41. Welcome to PF !

(Check the rules of this Forum regarding the use of bold face type.)

The initial value for the vertical component of velocity is 24m/s. If the vertical component of acceleration is -1.6 m/s2, then it takes 15 sec. for the projectile to reach it's max altitude, and another 15 sec. to return to launch elevation.

How much longer does it take to fall an additional 230 meters --- or is it 320 meters?

Last edited by a moderator:
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?
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If the ball is thrown down you are correct.
If the ball is thrown up your teacher is correct.