Question

A projectile is fired horizontally from a gun that is
45.0 m above flat ground, emerging from the gun with a speed of
250 m/s. (a) How long does the projectile remain in the air? (b) At
what horizontal distance from the firing point does it strike the
ground? (c) What is the magnitude of the vertical component of its
velocity as it strikes the ground?

Answer 1

t=3 s
and then?

Answer 2

How t=3s ?

Answer 3

45=0+(1/2)(10)(t^2)

Answer 4

yes that's correct !

Answer 5

Now, we will have to calculate the horizontal distance covered in the same time i.e 3 seonds

Answer 6

Just multiply 250 with 3 i.e speed with time.

Answer 7

Got it ?

Answer 8

(b) 750 m

Answer 9

yes,
Now for the third part we will have to calculate the final vertical velocity.
u=0
s=45
a=9.8

Calculate v by using \(\sf v^2=u^2+2as\)

Answer 10

\(\sf v^2=0+2\times 9.8\times 45\\v^2=882\\v=\sqrt{882}=29.69 m/s\)

Answer 11

i got 30 m/s :) i used g=10m/s^2

Answer 12

anyway, thanks

Answer 13

It's a good practice to use g=10 [if not stated]

Answer 14

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