A projectile of mass 0.317 kg is shot from
a cannon, at height 6.7 m, as shown in the
figure, with an initial velocity vi having a
horizontal component of 6.4 m/s.
The projectile rises to a maximum height
of 3.2 m above the end of the cannon’s barrel
and strikes the ground a horizontal distance
x past the end of the cannon’s barrel.
Find the magnitude of the final velocity vector
when the particle impacts the ground?
Do I break them into their x and y components or what should I do?? Please help
a cannon, at height 6.7 m, as shown in the
figure, with an initial velocity vi having a
horizontal component of 6.4 m/s.
The projectile rises to a maximum height
of 3.2 m above the end of the cannon’s barrel
and strikes the ground a horizontal distance
x past the end of the cannon’s barrel.
Find the magnitude of the final velocity vector
when the particle impacts the ground?
Do I break them into their x and y components or what should I do?? Please help
-
horizontal component of 6.4 m/s = constant
let vertical component be u
vertical motion,
s = 3.2; v=0
v2 =u2 +2as
0 = u^2 - 2 *9.81 * 3.2
u = 7.924m/s
s = - 6.7;
v^2 = u^2 +2as
v^2 = 7.924^2 - 2 * 9.81 * (-6.7)
v = 13.94m/s
magnitude of the final velocity vector
when the particle impacts the ground
= sq root [ 13.94^2 + 7.924^2]
m/s
answer
let vertical component be u
vertical motion,
s = 3.2; v=0
v2 =u2 +2as
0 = u^2 - 2 *9.81 * 3.2
u = 7.924m/s
s = - 6.7;
v^2 = u^2 +2as
v^2 = 7.924^2 - 2 * 9.81 * (-6.7)
v = 13.94m/s
magnitude of the final velocity vector
when the particle impacts the ground
= sq root [ 13.94^2 + 7.924^2]
m/s
answer
-
Sorry I made a mistake
magnitude of the final velocity vector
when the particle impacts the ground
= sq root [ 13.94^2 + 6.4^2]
m/s
answer
magnitude of the final velocity vector
when the particle impacts the ground
= sq root [ 13.94^2 + 6.4^2]
m/s
answer
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