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

In summary: 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
  • #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 :)
 
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  • #2
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?
 
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  • #3
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?
------------------------
If the ball is thrown down you are correct.
If the ball is thrown up your teacher is correct.
 

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

What is projectile motion?

Projectile motion is the motion of an object through the air, under the influence of gravity. It occurs when an object is thrown or launched at an angle to the ground.

What factors affect projectile motion?

The factors that affect projectile motion include the initial velocity, angle of launch, air resistance, and gravity.

How do you calculate the range of a projectile?

To calculate the range of a projectile, you can use the equation R = (v^2 * sin(2θ)) / g, where R is the range, v is the initial velocity, θ is the angle of launch, and g is the acceleration due to gravity.

What is the maximum height of a projectile?

The maximum height of a projectile occurs at the highest point of its trajectory, where the vertical velocity is equal to zero. It can be calculated using the equation h = (v^2 * sin^2(θ)) / (2 * g), where h is the maximum height.

How does air resistance affect projectile motion?

Air resistance can affect projectile motion by slowing down the object and changing its trajectory. This can lead to a shorter range and a lower maximum height. However, for most practical situations, air resistance can be ignored as it has a minimal impact on the motion of the object.

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