## throws a stone horizontally with a velocity of 20ms-1

1. The problem statement, all variables and given/known data

1. Don Joriel stands on a vertical cliff edge throwing stones into the sea below. He throws a stone horizontally with a velocity of 20ms-1, 560 m above sea level.
a. How long does it take for the stone to hit the water from leaving Don Joriel’s hand?
Use g = 9.81 ms-2 and ignore air resistance?
b. Find the distance of the stone from the base of the cliff when it hits the water.

2. When Maggie applies the brakes of her car, the car slows uniformly from 15.00 m/s to 0.00 m/s in 2.50 s. How many meters before a stop sign must she apply her brakes?

2. Relevant equations

3. The attempt at a solution
b. 20.39 m
2. 18.75 m

Hope you can help me, if got the wrong answer please show me the solution... Thanks a lot...
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 Recognitions: Homework Help Show your calculations.
 Recognitions: Homework Help What rl.bhat said: if you don't show us what you tried or your reasoning, then we cannot know how to help you best.

## throws a stone horizontally with a velocity of 20ms-1

wait i'll show you when i got home.. thanks...
 for 1a. i use v = u + at, b. s = ut +1/2 at squared for 2. s = 1/2 (u +v)t are those correct?

 Quote by moimoi24 for 1a. i use v = u + at, b. s = ut +1/2 at squared for 2. s = 1/2 (u +v)t are those correct?
I don't see how you can use the equation in 1a without first knowing the final velocity of the ball (I.e the velocity of the ball prior to impact with the ground). From your answer, it seems you have assumed this to be 0, but this is not the final vertical component of velocity.
 For 1b, we are considering a horizontal distance. There is no acceleration of the ball in x direction so the eqn you were using simplifies to simply $s = v_{ox}t$. For 2, if I understand the wording of the question correctly, you are correct.

 Quote by CAF123 I don't see how you can use the equation in 1a without first knowing the final velocity of the ball (I.e the velocity of the ball prior to impact with the ground). From your answer, it seems you have assumed this to be 0, but this is not the final vertical component of velocity.
 Quote by CAF123 For 1b, we are considering a horizontal distance. There is no acceleration of the ball in x direction so the eqn you were using simplifies to simply $s = v_{ox}t$. For 2, if I understand the wording of the question correctly, you are correct.

Recognitions:
Homework Help
 Quote by moimoi24 for 1a. i use v = u + at, b. s = ut +1/2 at squared for 2. s = 1/2 (u +v)t are those correct?
Before you attempt this problem, please go through the projectile motion.

 Quote by moimoi24 what is the answer then? please show the solution...
We are not allowed to simply hand over solutions. For 1a, you can use the equation $s_y = v_{oy}t - \frac{1}{2}gt^2$ right away to find the time of descent. Alternatively, the final vertical component of velocity can be found first so you can then use $v_{fy} = v_{oy} -gt$.
Yet another way is to do all three problems via sketching v-t diagrams for each of the x and y components of velocity.
 Quote by CAF123 We are not allowed to simply hand over solutions. For 1a, you can use the equation $s_y = v_{oy}t - \frac{1}{2}gt^2$ right away to find the time of descent. Alternatively, the final vertical component of velocity can be found first so you can then use $v_{fy} = v_{oy} -gt$. Yet another way is to do all three problems via sketching v-t diagrams for each of the x and y components of velocity.
Ok, for 1a, initial vertical component of velocity is 0, right? This is because the ball is projected horizontally. So in our equation, $v_{oy} =0.$ The eqn therefore reduces to $s_y = \frac{1}{2}at^2.$ If you define downwards as negative, then the ball will fall downwards 500m so $s_y = -500m.$ Think of the ball moving 500 units down the y axis (I.e starts at y =0 and finishes at y = -500). Now, if downwards is negative, a =-g and so our final equation is $-s_y = -\frac{1}{2}gt^2$ Rearrange for t and you are done with 1a.