An object is thrown upward from a height of 50 ft with an initial velocity of 48 ft per sec. The height in?

Question

cont.
feet after t seconds is given by h=-16t^2+48t+50
A) find the height of the object 3 sec after it is thrown upward.
B) Find the maximum height of the object and the time it takes the object to reach this height

Answer 1

h=-16t^2+48t+50
for first part put t=3
h=-16*9+48*3+50
solve and subtract 50 to find height above projection
for second part diff the eq. to get velocity
dh/dt=-32t+48
at max height this velocity is zero
0=-32t+48
t=3/2
put this in given eq. to find height

Answer 2

whats up Hardrockangel19! a) d = do + vt + 0.5at^2, do=5', v=30ft/s, a=-g the place g is the acceleration by way of gravity (9.8m/s - convert to English gadgets) consequently: d = do + vt - 0.5gt^2 b) set: do + vt - 0.5gt^2 = 0, and take the by-product and remedy for t. consequently t=v/g. plug that into d = do + vt - 0.5gt^2 and remedy for d c) from b: t=v/g d) using d = do + vt - 0.5gt^2, remedy for t with d=0 wish this enables! digital mail or IM me whether it is not sparkling.