after falling for 2 s ... the impact velocity is .... 2.00 s * g = 18.6 m/s
the average velocity is half of the impact velocity ... 9.81 m/s
the 2 makes things fairly simple
A rock is dropped from a bridge into the water below. If it falls for 2.00 seconds, what is its average velocity? What is its instantaneous velocity?
4 answers
2g = 19.6
a = -9.81 m/s^2
a is acceleration due to gravity
v = Vi - 9.81 t
v is velocity, Vi is initial velocity, zero if dropped
so here v = -9.81 t m/s
at t = 2 seconds, v = 9.81*2 = 19.62 meters/second at ground
is is linear in t and v goes from 0 to 19.62 in 2 seconds so
v average = (0 + 19.62)/2 = 9.81 m/s
To go on with problem:
h = Hi + Vi t - (9.81/2 )t^2
h is height in meters, Hi is initial height. t is 2 seconds
so here 0 = Hi + 0 - 4.9*2^2
so
0 = Hi - 4.9 * 4
Hi = 19.6 meters initial height
a is acceleration due to gravity
v = Vi - 9.81 t
v is velocity, Vi is initial velocity, zero if dropped
so here v = -9.81 t m/s
at t = 2 seconds, v = 9.81*2 = 19.62 meters/second at ground
is is linear in t and v goes from 0 to 19.62 in 2 seconds so
v average = (0 + 19.62)/2 = 9.81 m/s
To go on with problem:
h = Hi + Vi t - (9.81/2 )t^2
h is height in meters, Hi is initial height. t is 2 seconds
so here 0 = Hi + 0 - 4.9*2^2
so
0 = Hi - 4.9 * 4
Hi = 19.6 meters initial height
oops ... math error
0 answers