A projectile is fired at an upward angle from the top of a 105 m cliff with a speed of 185 m/s. What will be its sped when it strikes the ground below?(Use conservation of energy and neglect air resistance.)
2007-10-30 09:11:41 · 3 answers · asked by matt c 1 in Science & Mathematics ➔ Physics
Conservation of energy?
OK
Total energy = potential energy of the projectile before it leave s the gun plus kinetic energy of just fired projectile.
Et=Ke+Pe = 0.5mV^2 +mgh
When it strikes the ground it will have kinetic energy Kef equal to the total energy
Kef= Et=0.5 mVf^2
Vf= sqrt (2Et/m)
Vf=sqrt(2(0.5mV^2 +mgh)/m
Vf=sqrt(V^2 +2 gh)
Vf=sqrt(185 ^2 + 2x 9.81x 105 )=
Vf=190 m/s
2007-10-30 09:16:17 · answer #1 · answered by Edward 7 · 0⤊ 0⤋
energy is conserved
energy before = energy after
initially, the projecttile has both Potential energy and Kinetic energy, when it hits the ground, it's all Kinetic energy
PEo + KEo = KEf
mgh + .5mv^2 = .5mv^2
the mass cancel out
gh + .5v^2 = .5v^2
plug in chunks
9.8(105) + .5(185)^2 = .5v^2
solve for v and you'll get v = 190.48 m/s
2007-10-30 09:24:08 · answer #2 · answered by Anonymous · 0⤊ 0⤋
about 190 m/s
2007-10-30 09:18:18 · answer #3 · answered by Lola 2 · 0⤊ 0⤋