a ball is thrown upward with velocity 100m/sec. find velocity after travelling 180m from ground.?
Submitted : 20180614 06:57:08 Popularity:
Tags: upward velocity ball thrown travelling
Let g = 10 m / s²
v ² = u ²  2 g s
v ² = 100 ²  20 x 180
v ² = 100 ²  3600
v ² = 100 [ 100  36 ]
v² = 100 x 64
v = 80 m/sec
K
Let g = 10 m / s²
v ² = u ²  2 g s
v ² = 100 ²  20 x 180
v ² = 100 ²  3600
v ² = 100 [ 100  36 ]
v² = 100 x 64
v = 80 m/sec
KE = 1/2 (m*v^2) = 1/2 (1 x 100^2) = 5,000J.
PE = (mgh) = (1 x 9.8 x 180) = 1,764J.
(5,000 1,764) = 3,236 J.
V = sqrt.(2KE/m) = sqrt.(6,472/1)= 80.45m/sec.
This is actually V down, but is equal to V up at the same height.
So we have to assume that somehow that ball is thrown from ground level, h = 0.
From the conservation of energy we have TE = PE + KE = constant. So the initial total energy TE = 1/2 mU^2 and the total energy at h = 180 m is TE = 1/2 mv^2 + mgh.
But the two TE are equal; so we can write 1/2 mU^2 = 1/2 mv^2 + mgh and solve for v = sqrt(U^2  2gh) = sqrt(100^2  2*9.8*180) = 8.0449E+01 m/s when h = 180 m. ANS.
..,.
use this
Vf^2 = Vi^2 + 2 g d
Vf^2 = 100^2 + 2(9.81)(180) = 6468.4
Vf = 80.4 m/s (round appropriately)
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