There are several ways to solve this. An elegant way is using conservation of energy: If you neglect air resistance, after dropping 30 meters, all the potential energy is converted to kinetic energy. So, just calculate the potential energy at the top, assume the kinetic energy at the bottom is the same value, and solve the kinetic energy equation for speed.
velocity = acceleration x time where acceleration is 9.8 m/s/s (32.2 ft/s/s). This is 9.8 x 2.5 = 24.5 m/s (80 ft/sec) , or 55 mph.
Approx 24.5 metres/second.
39 m\s downward
The speed is 44.4... repeating metres per second.
They will both have the same speed because the will hit the ground at the same time, due to vertical velocity.
Simply use the expression v = gt g = 9.8 m/s^2 and t given as 4.5 s So velocity with which the penny hits the ground will be 44.1 m/s
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.
39 m\s downward
The speed is 44.4... repeating metres per second.
If the ball was dropped from a roof and hit the ground 3.03 seconds later, then when it hit the groundits velocity was 29.694 meters (97.42 feet) per second (rounded) downward.
They will both have the same speed because the will hit the ground at the same time, due to vertical velocity.
Simply use the expression v = gt g = 9.8 m/s^2 and t given as 4.5 s So velocity with which the penny hits the ground will be 44.1 m/s
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.
After 3.5 seconds of free-fall on or near the surface of the Earth, (ignoring effectsof air resistance), the vertical speed of an object starting from rest isg T = 3.5 g = 3.5 x 9.8 = 34.3 meters per second.With no initial horizontal component, the direction of such an object's velocitywhen it hits the ground is straight down.
The initial velocity of a dropped ball is zero in the y (up-down) direction. After it is dropped gravity causes an acceleration, which causes the velocity to increase. F = ma, The acceleration due to gravity creates a force on the mass of the ball.
Let v be the velocity when the ball is at 640 feets going downwards v = 48 feet /sec let the velocity with which it reaches the ground be u then, u2=v2+2gh g = acc due t ogravity in feet/sq.sec h = 640 feet the time taken to reach the ground = time to return to 640ft + the time to fall from there Time taken to get to the ground is 8 seconds. Final velocity is 208 feet per second downwards
In two seconds of fall, the speed increases 19.6 meters (64.4 feet) per second. The magnitude of velocity increases by that amount, while the direction of velocity doesn't change.
The volleyball will NOT hit the ground with greater anything. Assuming that the soccer ball is the same spherical diameter and greater mass than the volleyball it will hit the ground with greater velocity and greater impact.
Disregarding air resistance, what is the speed of a ball dropped from 12 feet just before it hits the ground? (Use 1 ft = 0.30 m, and use g = 9.8 m/s2.)