This is a relatively easy problem that is confused by two separate velocities.
When working with vertical velocity, Vy, you ignore what the horizontal velocity, Vx, is doing, and visa versa.
For A you can use this equation to solve for time until velocity is 0, which would be when the object is at its apex, and then multiply it by 2 to get the entire time it is in the air.
t=(0-Vi)/a
t=(0-49)/-9.8 <--- notice we only use the vertical component of the velocity for this, the horizontal does not matter. notice also that the acceleration from gravity is negative because it is in the opposite direction of our positive vertical velocity.
t=-49/-9.8=5. Time until velocity=0 at the top of the parabola is 5 seconds. Since no other force is acting on it you can multiply this by 2 to equal the entire time the projectile in flight, 10 seconds.
Now we have the entire time the object is moving vertically, this will also be the total time the object is moving horizontally. We have the speed that the object is moving horizontally, Vx=60m/s. Therefore in 10 seconds of travel the object will travel 600 meters horizontally.
Aprojectile Is At Such An Angle That
The Vertical Component Its Volocity
Is49m\s The Horizontal Component
Of Its Volocity Is 60m\s.
A,how Long Does The Projectile
Remain In Air?
B,what Horizontal Distance Does
It Travel?
2 answers
Xo = 60 m/s.
Yo = 49 m/s.
Vo = 60 + 49i = 77.5m/s[39.2o].
A. Y = Yo + g*Tr = 0.
49 + (-9.8)Tr = 0,
Tr = 5 s. = Rise time.
Tf = Tr = 5 s. = Fall time.
Tr+Tf = 5+5 = 10 s. = Time in air.
B. d = Xo*(Tr+Tf) = 60*10 = 600 m.
Yo = 49 m/s.
Vo = 60 + 49i = 77.5m/s[39.2o].
A. Y = Yo + g*Tr = 0.
49 + (-9.8)Tr = 0,
Tr = 5 s. = Rise time.
Tf = Tr = 5 s. = Fall time.
Tr+Tf = 5+5 = 10 s. = Time in air.
B. d = Xo*(Tr+Tf) = 60*10 = 600 m.
1 answer