u = 10 m/s forever
v = -gt = -9.81 (3) = - 29.4 m/s
check if it hit the ground
d = (1/2) g t^2 = 4.9 * 9 = 44.1 so still in the air
speed = sqrt (10^2 + 29.4^2)
a body is projected horizontally from height of 78.4 m with a velocity 10 m/s . its velocity after 3 sec
2 answers
First, check that it hasn't yet hit the ground.
Finding it has not, then get the resultant of the horizontal and vertical velocities
√(10^2 + (9.8*3)^2)
at an angle of 71° below the horizontal
Finding it has not, then get the resultant of the horizontal and vertical velocities
√(10^2 + (9.8*3)^2)
at an angle of 71° below the horizontal