Projectile motion at top of cliff with angle

In summary, the conversation discusses a projectile problem involving a ball thrown with an initial speed of 30 m/s at an angle of 45° from a height of 11 m. The equation Yf = Yo + Vot - (1/2)gt^2 is used, with the known values of Vo = 30sin45 = 21.213 m/s and g = -9.81. The person is confused about what values to use for Yf and Yo and how to manipulate the equation to find t. They suggest a possible solution of setting the equation equal to 0, but acknowledge that it may not be correct. The expert summarizer suggests solving the quadratic equation or using the vertical velocity to find the time
  • #1
aemaem0116
1
0
hi. i have this projectile problem and i can't seem to get this one question right.

A ball is thrown with an initial speed of 30 m/s at an angle of 45°.The ball is thrown from a height of 11 m and lands on the ground.

i know the equation i have to use is:

Yf = Yo + Vot - (1/2)gt^2

i know my Vo = 30sin45 = 21.213 m/s
g = -9.81
i know the given Y is 11 m but I am really confused about what should be my Yf and Yo. Most importantly, how can I manipulate the equation to find t??

what i thought it could be is like this...but i know its not right:

0 = 11 + (21.213)t - (.5)(-9.81)t^2
-11 = 21.213t - 4.905t^2

and that's about the farthest i could get...which isn't very far
 
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  • #2
aemaem0116 said:
hi. i have this projectile problem and i can't seem to get this one question right.

A ball is thrown with an initial speed of 30 m/s at an angle of 45°.The ball is thrown from a height of 11 m and lands on the ground.

i know the equation i have to use is:

Yf = Yo + Vot - (1/2)gt^2

i know my Vo = 30sin45 = 21.213 m/s
g = -9.81
i know the given Y is 11 m but I am really confused about what should be my Yf and Yo. Most importantly, how can I manipulate the equation to find t??

what i thought it could be is like this...but i know its not right:

0 = 11 + (21.213)t - (.5)(-9.81)t^2
-11 = 21.213t - 4.905t^2

and that's about the farthest i could get...which isn't very far

Looks about OK. (Lose the - in front of the -9.8 as you did when you rearranged.)

That gives you the time until it hits the ground.

Simply solve the quadratic.

Alternatively you could use the 21.213 as the vertical velocity and figure the time and height of max height and then use x = 1/2*g*t2 to determine the time to fall and add the two times together.

Either way works.
 
  • #3
.

Hi there,

Based on the information given, it seems like you are on the right track with using the equation Yf = Yo + Vot - (1/2)gt^2. In this case, Yf refers to the final height, which we know is 0 since the ball lands on the ground. Yo refers to the initial height, which is 11 m. So, our equation becomes:

0 = 11 + (21.213)t - (1/2)(-9.81)t^2

Next, we need to solve for t, which is the time it takes for the ball to reach the ground. To do this, we can rearrange the equation to get it in the form of a quadratic equation:

-4.905t^2 + 21.213t + 11 = 0

We can then use the quadratic formula to solve for t:

t = (-b ± √(b^2 - 4ac)) / 2a

In this case, a = -4.905, b = 21.213, and c = 11. Plugging these values into the formula, we get two solutions for t: t = 2.024 seconds or t = 4.519 seconds.

However, since we are looking for the time it takes for the ball to reach the ground, we can discard the t = 4.519 seconds solution, as this is the time it would take for the ball to reach its maximum height and then fall back down to the ground. So, the correct answer is t = 2.024 seconds.

I hope this helps and clarifies any confusion you had. Keep in mind, when solving projectile motion problems, it's important to carefully consider what each variable represents and to use the correct formula for the situation. Best of luck with your problem!
 

Related to Projectile motion at top of cliff with angle

1. How is the angle of the cliff related to the projectile motion at the top?

The angle of the cliff affects the initial velocity and trajectory of the projectile at the top. The steeper the angle, the greater the initial velocity and the shorter the horizontal distance traveled by the projectile.

2. What is the maximum height reached by the projectile at the top of the cliff?

The maximum height reached by the projectile at the top of the cliff can be calculated using the formula h = v2sin2θ / 2g, where h is the maximum height, v is the initial velocity, θ is the angle of the cliff, and g is the acceleration due to gravity.

3. How does air resistance affect the projectile motion at the top of the cliff?

Air resistance can affect the trajectory of the projectile at the top of the cliff, causing it to deviate from the expected path. However, the effect of air resistance is usually negligible for most real-life situations.

4. Can the horizontal distance traveled by the projectile at the top of the cliff be greater than the height of the cliff?

Yes, it is possible for the horizontal distance traveled by the projectile to be greater than the height of the cliff if the angle of the cliff is shallow and the initial velocity is high.

5. How does the mass of the projectile affect its motion at the top of the cliff?

The mass of the projectile does not affect its motion at the top of the cliff, as long as the air resistance is negligible. This is because the acceleration due to gravity is constant for all objects, regardless of their mass.

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