Physics Question Doubt : projectiles

[aiyla]

The Question says :
Q) A baseball is hit at ground level the ball reaches its max.height above the ground level 3s after being hit. Then 2.5s after reaching the max height, the ball barely clears a fence that is 97.5m from where it was hit.

a) What is the max height ?
b) How high is the fence ?
c) how far beyond the fence does the ball strike ?


thanks

[vanibharutham]

a) Assume the highest point is at x

Time taken to reach x = 3 seconds
At x, v = 0 m/s
And the only force acting will be downwards from the acceleration, therefore a = -9.8

v = u + at
0 = u + (-9.8 x 3 )
u = 29.4 m/s

This is just the VERTICAL INITIAL VELOCITY

now

v² = u² + 2as

Substitute
and you get
s = 44.1 meters


PLEASE CHECK THIS not 100% sure

[vanibharutham]

b) Now the ball is at x, and is going downwards

So,

a = 9.8 m/s²
u = 0 m/s
v = ?
t = 2.5 s

v = u + at
v = (0) + (9.8 x 2.5)
v = 24.5 m/s

Now
v² = u² + 2as
(24.5)² = 0² + 2(9.8)s
Giving,

s = 30.625 m

this is how much the ball has travelled in the 2.5 seconds,

Therefore the height of the fence =

Max height - 30.625

44.1 - 30.625  = 13.475 m


Do u have the answers to these, or could astar check em

[vanibharutham]

c) First, we need to calculate the time it took to reach the ground past the fence...

s = ut + ½at²
13.475 = 24.5t + 4.9t²
4.9t² + 24.5t - 13.475 = 0
If you use the quadratic equation, you get

t = 0.5 seconds and t = -5.5 seconds

Discard the negative value...
It takes 0.5 seconds from the top of the fence to land to the ground... Time of flight = 6 seconds



NOW, we will consider the horizontal components only

We know that it took 5.5 seconds to cover 97.5 metres
a = 0 ms² (BECAUSE IT IS HORIZONTAL)
u = ?
s = 97.5m
t = 5.5s

s = ut + ½at²
97.5 = 5.5u

Giving,

u = 17.72727272727272 m/s

If it travelled for 6 seconds,

s = 17.72727272727272 x 6
s = 106.36363636 m

Distance travelled past fence = 106.363636363636 - 97.5 = 8.864 metres (4 sig. fig)

Hope that helped :S

and please check

[nid404]

a) v=u+at
0=u-gt(-9.81X3)
u=29.4
This is vertical velocity
v2=u2+2as
therfore
02=29.4²+2X-9.81Xs(h)
it gives h=44.05m

i get the same thing.....

b)

v=u +at
=0+(2X-9.81h)
v=24.525
v2=u2+2as
s=30.66

distance travelled in 2.5s is 30.66m
u know s=vcosthetaXtime (vectors)
30.66=24.525costhetaX2.5
cos theta=0.5..
theta=59.6
substitute in
h=vsinthetaXtime - 0.5X-9.81X time²
h=22.3 m
I'm not sure....plz check

If you can verify the 2nd part...i will move on to the third one...im not very confident abt my answer

[aiyla]

thanks every one, i dont really have the answers .. but your ways seams correct .
thanks alot this helped

[nid404]

thanks every one, i dont really have the answers .. but your ways seams correct .
thanks alot this helped

oh ok....i will confirm with my engineering professor and let you knw by tom

[aiyla]

thanks alot nid =D