An object is observed to fall from a bridge, striking the water below 2.9 s later.?

Question

An object is observed to fall from a bridge, striking the water below 2.9 s later.

(a) With what speed did it strike the water? (m/s)

(b) What was the average speed during the fall? (m/s)

(c) How high is the bridge? (m)

Answer 1

I won't do you homework for you, but I'll give you hints.

v = a * t

We are given t is 2.9 and you should know a is 9.8 (gravity) so that's how you find the first answer.

d = (1/2) a * t^2

We know t is 2.9 and a is 9.8 (gravity) so that gives you the third answer.

d = v * t

We just calculated d and we were given t so substitute in those numbers and solve for v, that's your average speed.

Answer 2

You can run this backwards:
c) x = 4.9 * tf^2, where tf is 2.9 seconds.
a) v(tf) = 9.8 * tf, where t is as before
b) This is a bit tricky, since the definition of "average speed" is NOT 0+v(tf)/2. It is
(1/tf) S(v(t)dt . But the integral is x, so the average speed is x/tf.

Answer 3

u=0
a=10m/s^2, t=2.9s, v=?
v=u+at
=0+10*2.9
=29m/s
average speed= (u+v)/2
=(0+29)/2
=14.5m/s
c) s=ut+1/2 at^2
= 0+1/2 *10 * 2.9^2
= 5*8.41
=42.05m