Velocity/motion problem?

Question

A body is projected nearly vertically upwards with a speed of 34.3 m/s from a cliff 147 m high. How long will it be before it reaches the ground below and what will be its speed just before it hits the ground?

I've tried a lot of different things, drawn many a diagram, and i can't seem to get the answers. The answers are 10s and 63.7m/s.

Any help with this one is appreciated. :)

Answer 1

When thrown up V= -34.3 m/sec

It will go up a distance (reaches 0 velocity) and then start dropping

So using

V^2 –U^2= 2as

Where a= 9.81 m/sec^2 and u=0

Distance is 60 mtrs.

So it will go up 60 mtrs and then drop 60+147 mts before hitting the ground.

Time taken for up travel is

V=u +at

So t= 3.5 secs.

So it will travel up for 3.5 seconds and then 3.5 seconds to reach original starting point.

Now for findings final velocity use

V^2 – U^2= 2as

Where u= 34.3 m/sec and s= 147 and a is acceleration due to gravity which is 9.81

Answer is 63.72 m/sec

time taken is v=u +at

63.72= 34.3 + 9.81 *t

t=3 seconds

So total time of travel is 3.5 (up) plus 3.5 down to same level and 3 to 147 below which is 10 seconds.

Answer 2

Time until the maximum height the body attains.

Vf - Vi = at

0 - 34.3 = -10t

t=3.43s

And how high that is from the top of the cliff.

Vf^2 - Vi^2 = 2ad

0 - 34.3^2 = 2x-10d

d = 58.8m

Velocity of the body just before impact at the bottom of the cliff.

Vf^2 - Vi^2 = 2ad

Vf^2 - 0 = 2x-10x-205.8

Vf = -64m/s (fairly close to the answer, I rounded off)

Time taken from maximum height to bottom of cliff.

Vf - Vi = at

-64 - 0 = -10t

t = 6.4s

Total time taken.

3.43 + 6.4 = 9.83 (which isn't too bad)

If you want more accurate answers then use -9.81 instead of -10 for the acceleration due to gravity and don't round answers half way through like I did.