Projectile motion at top of cliff with angle

[aemaem0116]

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

 

[LowlyPion]

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.

 

[thomate1]

.

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!