2 Questions!

i dont have a clue how to do these questions :mad:

1) Tennis player hits ball vertically at speed of 10m/s how high does it go?

2)Stone is dropped from a window 5m high at what speed does it hit the ground?

:frown:

thanks in advance :smile:

Reply 1

Intelligentsia

I'll give you a hint. Use SUVAT equations. And use the fact that projectiles under free fall always have an acceleration or deceleration of g i.e. 9.8m/s2. And remember to use your signs carefully.

Reply 2

welwyn

or for the second one you could use the equation GPE = KE assuming no energy loss, then word out the speed from the KE

Reply 3

Gingerbread Man

SUVAT equations :confused:

Are they a level stuff? never heard of em could be coz we have bad teacher though!

welwyn i see what u mean by using GPE=KE but i dont know the mass so im kinda stuck :frown:

For both questions only way i know i could work em out is if i knew mass :redface:

Reply 4

Robob

You dont need the mass, 0.5mv^2 = mgh
0.5v^2 = gh
h = 0.5v^2 / g

rearrange differently for second question, as you can see the mass cancels

suvat equations are: v = u + at etc (have you seen that equation)

Reply 5

welwyn

if GPE = KE then the mass cancels from both sides.

and about 1) - use v^2 = u^2 + 2as
ie
the final velocity squared = the initial velocity squared + (2 x accn. of freefall x distance)
so
0 = 100 + (2 x -9.81 x s)

This is the easiest way, but I can't instantly think of a way with the mass.

welwyn

Reply 6

Gingerbread Man

For 1) i got 5-lol hope its correct :smile:

for 2) im still kinda confused :frown:

thanks again for help so far its REALLY been appreciated

Reply 7

Mr power

yer suvat equations are a level stuff
s= distance
v= velocity
u= initial velocity
a = acceleartion

Reply 8

Gregorian IV

anon234

i dont have a clue how to do these questions :mad:

1) Tennis player hits ball vertically at speed of 10m/s how high does it go?

2)Stone is dropped from a window 5m high at what speed does it hit the ground?

:frown:

thanks in advance :smile:

Use the equation v^2=u^2+2as

where v = final velocity
u = initial velocity
s = displacement
a = acceleration (9.81 m/s/s)

So

1) rearrange teh equation to make s the subject, u=10m/s, v = 0 m/s and a = 9.81 m/s/s

s = (v^2-u^2)/2a
= (0 - 100) / 19.62
= 5.1m (if you ignore the '-' sign)

2) use the normal equation v^2 = u^2 + 2as v = subject, u = 0 a = 9.81 s = 5

v^2 = 2*9.81*5
v^2 = 98.1
v = 9.9 m/s