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. No need for any calculations: 1. 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. 2. Find the time the ball takes to get to the highest point in its trajectory by using t = v0y/g. 3. Find the height of the ball at its highest point by using y = y0 + v0y – (1/2)gt2. 4. Find the time it takes to fall from the highest point of its trajectory to the ground by using t = (2y/g). ONLY PUT STEPS IN ORDER

Explanation / Answer

Ans:In order to solve you need to follow the speps from 3-2-4-1.

Explanation. Suppose u through the ball at a initial velocity V0 and its x & y- component is V0x & V0y respectively.Here Y0 is the height of the roof.Now there is no force along x-direction but there is gravitation along y-direction, which will try to pull the ball down. But V0x is unchanged.

a)time taken to reach the maximum height is t = V0y/g ;

b) the height of the ball at the heighest point is Y= Y0 + V0y*t - (1/2)*g*t2

c)the time it takes to fall from the highest point of its trajectory to the ground T = sqrt(2*Y/g);

d) the distance the ball lands from the building =V0x*(t +T)

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