A cricket jumps vertically upwards, and remains in the air for 1.6 s. At
what speed did it leave the ground?
v = Vi - g t
at the top, v = 0 and t = 1.6/2 = 0.8 s
g = 9.81 m/s^2 approximately
so
0 = Vi -9.81(0.8)
Vi = 9.81(0.8) meters/ scond
To calculate the speed at which the cricket left the ground, we can make use of the motion equation which relates velocity, time, and acceleration.
The equation we can use is:
v = u + at
Where:
v is the final velocity (in this case, the speed at which the cricket left the ground)
u is the initial velocity (which is zero if the cricket jumped from rest)
a is the acceleration (due to gravity)
t is the time the cricket remains in the air (1.6 s in this case)
Since the cricket jumps vertically upwards, the acceleration due to gravity will act against the motion. Therefore, the acceleration can be taken as -9.8 m/s^2 (negative because it acts in the opposite direction of the jump).
Substituting the values into the equation:
v = 0 + (-9.8 m/s^2)(1.6 s)
v = -15.68 m/s
The negative sign indicates that the velocity is directed in the opposite direction (downwards). However, since the question asks for the speed (magnitude of velocity), we can ignore the negative sign and take the absolute value:
Speed at which the cricket left the ground = |-15.68 m/s| = 15.68 m/s
Therefore, the cricket left the ground with a speed of approximately 15.68 m/s.