Find the height from which the ball was thrown

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SUMMARY

The discussion focuses on solving a physics problem involving projectile motion, specifically determining the height from which a ball is thrown. The ball is launched with an initial velocity of 8.0 m/s at a 20.0-degree angle below the horizontal and strikes the ground after 3.00 seconds. The correct approach to calculate the height involves using the formula h = (v^2 * sin^2(theta)) / (2 * g), but it is emphasized that this formula is applicable only when the initial height is zero. Participants are advised to utilize basic kinematics equations for accurate results.

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Homework Statement


A ball is tossed from an upper story window of a building. The ball is given an initial velocity of 8.0m/s at an angle of 20.0 below the horizontal. It strikes the ground 3.00s later.

a) How far horizontally from the base of the building does the ball strike the ground?
b) Find the height from which the ball was thrown
c)How long does it take the ball to reach a point 10.0 below the landing?


Homework Equations


for a. the equation is R=vxit=vicostheta)*2t
for b. h=v^2sin^2thetai/2*g
for c. I think its Rmax


The Attempt at a Solution


ive plugged in the values to the equations but i still don't get the correct answer
 
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Part a) should be straightforward. What is the initial horizontal velocity? Does the horizontal velocity change at all?

you're also given the time... so you can calculate the horizontal displacement.
 
innightmare;1432888 [h2 said:
Homework Equations[/h2]
for a. the equation is R=vxit=vicostheta)*2t

Why 2t?

for b. h=v^2sin^2thetai/2*g

They're not asking for the height the object reaches (which you won't get from this equation anyway... this equation only works when the initial height is 0)... they're asking for the height from which it is thrown.

Don't try to use those formulas... stick to the basic kinematics formulas. Keep it simple.
 

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