A ball thrown into the air from a building 60 mtrs high,travels along a path given by h(x)= 60+x-x2/60,
where x is the horizontal distance from the building and h(x) is the height of the ball.What is the maximum height the ball will reach?

Dear Student,
Please find below the solution to the asked query:

hx=60+x-x260Differentiate with respect to x, we get:h'x=ddx60+x-x260h'x=0+1-2x60h'x=1-x30For maixma or minim, h'x=01-x30=0x30=1x=30 metresh'x=1-x30Again differentiating with respect to x, we get:h"x=-130 which is negative, hence hx will be maximum at x=30hx=60+x-x260h30=60+30-90060=90-15=75 mHeight of building is 60 m, hence maximum height reached by ball fromground is given by:hmax=60+75=135 m 

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