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Projectile motion at top of cliff with angle...

by aemaem0116
Tags: angle, cliff, motion, projectile
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aemaem0116
#1
Jan21-09, 10:17 PM
P: 1
hi. i have this projectile problem and i cant 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 im 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 thats about the farthest i could get...which isnt very far
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LowlyPion
#2
Jan21-09, 11:34 PM
HW Helper
P: 5,341
Quote Quote by aemaem0116 View Post
hi. i have this projectile problem and i cant 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 im 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 thats about the farthest i could get...which isnt 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.


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