Question

A shotputter throws the shot with an initial speed of 15.6 m/s at a 30.0 degree angle to the horizontal. Calculate the horizontal distance traveled by the shot if it leaves the athlete's hand at a height of 2.20m above the ground.

So I got that: v0= 15.6 m/s, vx0= 15.6 cos(30)=13.5,
vy0=15.6sin(30)=7.8, y=2.20m,

For the first part I did: y=y0+v0y t -gt^2
-> 0=-4.9t^2+7.8t+2.20
t= .366, 1.23....I chose 1.23seconds for this part.

Then:
x=0, t=1.23, x=?, v0x=13.5
x=x0+v0xt
x=15.6 m/s(1.23s)
I got 19.188 But the according to the online hw. The answer is wrong. Opinions? plz

Answer 1

Did I use the wrong formulas? Please help!

Answer 2

is the answer 34.13??
If yes i'll explain the method..

Answer 3

I got 9m

is it correct ?

Answer 4

I got 24.79m.
Message me if this is right & i'll tell you how i got it. Your method is right, though.

Answer 5

The answer ended up being 24.79 m. Hopefully there will be a solution posted as to how the answer was approached. Thank you so much everyone for all your help. This question was a little big more difficult than I thought. Thank you guys soo much!!!

Answer 6

As I told you, your method is absolutely right. It's just that the roots of the quad equation for time are t=-0.24 or t=1.84
take t=1.84
And now in your ques, you have written:
x=15.6 m/s(1.23s)
It should actually be
x=15.6cos(30) * 1.86
x=24.79m

Answer 7

@rajathsbhat which formula did you use...
dont you have to use 2 formulas to get the Range till it reaches 2.20 height, and a different formula to reach the ground?

Answer 8

if you know the time the shot-put spend in the air, you can simply calculate how far it has traveled by using
\[x=v_{x}t\]
because the horizontal velocity of the ball never changes.

Answer 9

any doubts @Anas.P ?

Answer 10

k i get it that if we know the speed and time w can find out the distance.
but what does x represent in your formula?

Answer 11

it represents the horizontal distance traveled.