- #1
brandon26
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A particle is projected on a horizontal ground and moves freely under gravity.
The horizontal and vertical components of the initial velocity are 2u and u ms^1 respectively.
The ball lands a distance 80m from the point of projection.
Show that u=14
This is what I did:
Taking motion in the horizontal plane------
2u = 80/t
Taking motion in the vertical plane-----
displacement (s) =0
acceleration (a)= -9.8
initial velocity (u) = u
time of motion (t) = t
therefore using s=ut +(1/2)at^2
I get: 4.9t=u
substituting this into the first equation should give me the correct value of u, but it doesnt. Can someone tell me where i went wrong?
The horizontal and vertical components of the initial velocity are 2u and u ms^1 respectively.
The ball lands a distance 80m from the point of projection.
Show that u=14
This is what I did:
Taking motion in the horizontal plane------
2u = 80/t
Taking motion in the vertical plane-----
displacement (s) =0
acceleration (a)= -9.8
initial velocity (u) = u
time of motion (t) = t
therefore using s=ut +(1/2)at^2
I get: 4.9t=u
substituting this into the first equation should give me the correct value of u, but it doesnt. Can someone tell me where i went wrong?