You throw a ball off a roof at some initial velocity directed upward and away fr

Question

You throw a ball off a roof at some initial velocity directed upward and away from the biulding. You'd like to calculate where it will land, but hate solving quadratic equations. There is another way you could solve the problem. Put following steps in the correct order to solve the problem while avoiding any quadratic headaches.

(See the hint if you need some guidance.)

Question 4 options:

1234

Find the time the ball takes to get to the highest point in its trajectory by usingt = v0y/g.

1234

Find the distance the ball lands from the building by adding the time it takes to get to the highest point to the time it takes to fall and multiplying thus sum by v0x.

1234

Find the time it takes to fall from the highest point of its trajectory to the ground by using t = (2y/g).

1234

Find the height of the ball at its highest point by using y = y0 + v0y – (1/2)gt2.

Save

1234

Find the time the ball takes to get to the highest point in its trajectory by usingt = v0y/g.

1234

Find the distance the ball lands from the building by adding the time it takes to get to the highest point to the time it takes to fall and multiplying thus sum by v0x.

1234

Find the time it takes to fall from the highest point of its trajectory to the ground by using t = (2y/g).

1234

Find the height of the ball at its highest point by using y = y0 + v0y – (1/2)gt2.

Explanation / Answer

Inorder to reduce the complexity of quadratic equation, need to solve these in following steps:

1) Time to reach to highest point from groung with initial velocity by t1 = uy/g

2) Height of the ball at highest point where vy is zero.

3) Time to reach to ground from highest level by t2= (2y/g)1/2

4)Total distance travelled is done by adding both the time and multipying with ux

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