time to hit the ground: 16t^2 = 15
horizontal distance: 80t
A ball is thrown straight out at 80 feet per second from an upstairs window that's 15 feet off the ground. Find the ball's horizontal distance from the window at the moment it strikes the ground.
2 answers
Yes but if you are supposed to do it from first principles using calculus then:
Vertical problem:
Force = mass * acceleration (Newton #2)
Force is gravity, m g down
g is acceleration of gravity, about 32 ft/s^2 down
so
m a = -32 m
a = -32 ft/s^2
then if h is distance above ground
then d^2 h/dt^2 = a = -32
so
dh/dt = initial speed up call it Vi - 32 t
h = initial height +Vi t - 16t^2
here initial height is 15 ft, we want when h = 0, at ground and Vi = 0
0 = 15 + 0 t - 16 t^2
16 t^2 = 15 as oobleck told you
solve for t
then
Horizontal problem:
no horizontal force so no horizontal acceleration
distance = 80 t
Vertical problem:
Force = mass * acceleration (Newton #2)
Force is gravity, m g down
g is acceleration of gravity, about 32 ft/s^2 down
so
m a = -32 m
a = -32 ft/s^2
then if h is distance above ground
then d^2 h/dt^2 = a = -32
so
dh/dt = initial speed up call it Vi - 32 t
h = initial height +Vi t - 16t^2
here initial height is 15 ft, we want when h = 0, at ground and Vi = 0
0 = 15 + 0 t - 16 t^2
16 t^2 = 15 as oobleck told you
solve for t
then
Horizontal problem:
no horizontal force so no horizontal acceleration
distance = 80 t
1 answer