Question
What is the velocity of a projectile with initial velocity 2m/s at an angle of 30degree to horizontal at the time 0.2s,0.5s,0.8s.Find the maximum height attained by projectile,total horinzontal distance travelled,seconds the projectile remains in air?what is the velocity of touching the ground(hit velocity)of a fall dropped from a height of 3m?
Answers
Vo = 2 m/s[30o].
Xo = 2*Cos30 = 1.73 m/s.
Yo = 2*sin30 = 1.0 m/s.
Y = Yo + g*Tr.
Y = 1 -9.8*Tr = 0,
-9.8Tr = -1,
Tr = 0.102 s. = Rise time or time to reach max ht.
Tf = Tr = 0.102 s. = Fall time.
Tr+Tf = 0.102 + 0.102 = 0.204 s. = Time in air.
Y = 1 - 9.8*0.2 = -0.96 m/s.
Y = 1 - 9.8*0.5 = -3.9 m/s.
Y = 1 - 9.8*0.8 = -6.84 m/s.
Negative velocities means the projectile is falling.
Y^2 = Yo^2 + 2g*h. Y = 0, Yo = 1.0 m/s, g = -9.8 m/s^2, h = ?.
Dx = Vo^2*sin(2A)/g. A = 30o, g = 9.8 m/s^2.
V = Vo = 2m/s[30o] = Velocity at which the projectile strikes gnd.
V^2 = Vo^2 + 2g*h. Vo = 0, g = 9.8 m/s^2, h = 3 m, V = ?.
Xo = 2*Cos30 = 1.73 m/s.
Yo = 2*sin30 = 1.0 m/s.
Y = Yo + g*Tr.
Y = 1 -9.8*Tr = 0,
-9.8Tr = -1,
Tr = 0.102 s. = Rise time or time to reach max ht.
Tf = Tr = 0.102 s. = Fall time.
Tr+Tf = 0.102 + 0.102 = 0.204 s. = Time in air.
Y = 1 - 9.8*0.2 = -0.96 m/s.
Y = 1 - 9.8*0.5 = -3.9 m/s.
Y = 1 - 9.8*0.8 = -6.84 m/s.
Negative velocities means the projectile is falling.
Y^2 = Yo^2 + 2g*h. Y = 0, Yo = 1.0 m/s, g = -9.8 m/s^2, h = ?.
Dx = Vo^2*sin(2A)/g. A = 30o, g = 9.8 m/s^2.
V = Vo = 2m/s[30o] = Velocity at which the projectile strikes gnd.
V^2 = Vo^2 + 2g*h. Vo = 0, g = 9.8 m/s^2, h = 3 m, V = ?.