Assuming you throw the rock horizontally off the cliff it drops down at the acceletrtion of gravity.
height= 1/2 gt^2
With g = 9.8 m/sec and t = 5 seconds we have
height = (1/2) (9.8)(5)(5) = 122.5 meters
notice it has nothing todo with the 50 meter distance, which depends on the horizontal velocity.
10 m/s
The ball was thrown horizontally at 10 meters per sec, and the thrower's arm was 78.4 meters above the base of the cliff.
if the bal is thrown by making 45 degree angles. with the ground..it will travel maximum distance...
64 METERSA+
The equation for vertical motion is y = v0t + .5at2. y is vertical displacement v0 is initial vertical velocity a is acceleration (in meters, normal gravitational acceleration is about -9.8 m/s/s, assuming positive y is upward displacement and negative y is downward displacement)
10 m/s
The ball was thrown horizontally at 10 meters per sec, and the thrower's arm was 78.4 meters above the base of the cliff.
22.35294117647059
Answer: 3 seconds
If it was thrown horizontally or dropped, and hit the ground 3.03 seconds later, then it hit the ground moving at a speed of 29.694 meters (97.42-ft) per second. If it was tossed at any angle not horizontal, and hit the ground 3.03 seconds later, we need to know the direction it was launched, in order to calculate the speed with which it hit the ground.
if the bal is thrown by making 45 degree angles. with the ground..it will travel maximum distance...
the distance it travels before falling to the ground
64 METERSA+
In the case of constant velocity (or speed), velocity = distance / time.
The equation for vertical motion is y = v0t + .5at2. y is vertical displacement v0 is initial vertical velocity a is acceleration (in meters, normal gravitational acceleration is about -9.8 m/s/s, assuming positive y is upward displacement and negative y is downward displacement)
1000000 m
Answer: 44 meters