Latest self Aptitude Question SOLUTION: At time t= 0, a projectile was fired upward from an initial height of 10 feet. Its height after t seconds is given by the function h(t)=p-10(q-t)^2, where p and q are positive cons

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(#M40010049) SELF question Keep an EYE

At time t= 0, a projectile was fired upward from an initial height of 10 feet. Its height after t seconds is given by the function h(t)=p-10(q-t)^2, where p and q are positive constants. If the projectilereached a maximum height of 100 feet when t= 3, then what was the height, in feet, of the projectile when t= 4 ?

chaitanya (11 years ago)

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jyotish r  (11 years ago)

At time t= 0, a projectile was fired upward from an initial height of 10 feet.
so, p-10q^2=10

projectilereached a maximum height of 100 feet when t= 3
so p-10(q-3)^2=100

solving these eqns, we get,
p=100 and q=3
he height, in feet, of the projectile when t= 4 is h(3)=100-10*(3-4)^2=90 feet

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jomon v j  (11 years ago)

h(t)=p-10(q-t)²

velocity = h´(t)=20(q-t),whe body reaches maximum height velocity=0


therefore q=t=3

also 100=p-10(3-3) so p=100

h(t)=100-(3-t)²

when t=4 h(t)=100-10=90

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