Question

The function H(t) = -16t2 + vt + s shows the height H (t), in feet, of a projectile launched vertically from s feet above the ground after t seconds. The initial speed of the projectile is v feet per second.

Part A: The projectile was launched from a height of 90 feet with an initial velocity of 50 feet per second. Create an equation to find the time taken by the projectile to fall on the ground.
Part B: What is the maximum height that the projectile will reach? Show your work.
Part C: Another object moves in the air along the path of g(t) = 28 + 48.8t where g(t) is th

Answer 1

A. H(t)=-16t^2+v t+s
s=90
v=50 ft/sec
when the projectile is on the ground H=0
\[0=-16 t^2+50 t+90,find ~t\]

Answer 2

ok thanks

Answer 3

B.
again \[H=-16t^2+50t+90,\frac{ dH }{ dt }=-32 t+50\]
\[when~ the~ body ~is ~at ~maximum~ height \frac{ dH }{ dt }=0\]
find t and then H at this t.

Answer 4

this is to confusing for me :/

Answer 5

for part B initial velocity v=50 given
when the projectile is at maximum height its velocity becomes zero ,then it comes back to ground.

Answer 6

\[\frac{ dH }{ dt }\]
gives the velocity of projectile at any time t.

Answer 7

-32t+50=0
\[t=\frac{ 50 }{ 32 }=\frac{ 25 }{ 16 } \sec\]
find H at t=25/16 sec