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 ?

• 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
• 12 years agoHelpfull: Yes(2) No(1)
• 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
• 12 years agoHelpfull: Yes(0) No(1)