Ignoring air resistance, its horizontal speed is still 9 meters per second, its vertical speed is approx. 9.81 m/s, as the acceleration of gravity is 9.81 meters per second per second.
After a second, the ball will still have a horizontal velocity of 8 meters per second. It will also have a vertical velocity of 9.8 meters per second (Earth's acceleration is about 9.8 meters per square second). The combined speed (using the Law of Pythagoras) is about 12.65 meters per second.
t=(square root of)(2d/g)t=2(78.4)/9.8t=156.8/9.8t=(square root of)16t=4 seconds
Excluding external influences, on earth, you accelerate at 9.81 meters per second per second (1 Gravity).
If the ball ended 40 m from the cliff after 4 seconds, the original horizontal velocity was 10 m/s. You can calculate the height assuming standard gravity acceleration (g = 9.81 m/s2) and zero original vertical velocity, by calculating the final downward velocity vy2, vy2 = a x t = 9.81 m/s2 x 4 s = 39.24 m/s and from this the average velocity, vav = 39.24 m/s / 2 =19.62 m/s and multiply by 4 s, 19.62 m/s x 4 s = 78.48 m.
Assuming that the gun is fired parallel to the ground, the bullet will begin to fall the instant that it leaves the muzzle. The total fall will be 200 meters. You will need to calculate how long it will take an object to fall 200 meters (hint- about 9.753 meters per second per second- or 9.753 meters the first second, 19.50 meters during the second second, etc) THEN multiply the velocity of the bullet (643 meters/ second) by the number of seconds it is in flight. That will be the distance when it hits the ground.
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.
Answer: 44 meters
64 METERSA+
10 m/s
64 metersIf a ball is thrown horizontally at 20 m/s from the top of a cliff that is 50 meters high, the ball will strike the ground 64 m from the base of the cliff (20m/s x 3.2 s).
Answer: 3 seconds
After a second, the ball will still have a horizontal velocity of 8 meters per second. It will also have a vertical velocity of 9.8 meters per second (Earth's acceleration is about 9.8 meters per square second). The combined speed (using the Law of Pythagoras) is about 12.65 meters per second.
"3.2" or "3.20" please put all of that
An object thrown upward at an angle An object that's thrown horizontally off a cliff and allowed to fall
If it was thrown horizontally, it had an initial velocity of 10 meters/sec parallel to the ground. (It traveled 40 meters in 4 secs with no acceleration. x=vt) It also took 4 secs to travel vertically. It started with a vertical velocity of 0 m/s. Using x=v0 + (1/2) a t2 a = -g ( Acceleration due to gravity 9.8m/s2) x=0-(1/2)g*16 = -8 * 9.8 = -78.4 m It fell 78.4 meters before coming to a stop.
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.
the time it takes to get to ground is s = 1/2 at^2 where a is gravity acceleration of 9.8 m/s/s and s is 50 meters and t^2 is time squared.Solve for time t = 3.2 secondsThe horizontal distance from base of cliff is h = vt where h is distance and v is velocity = 20 m/sh = 20 x 3.2 = 64 meters