Best Answer

According to the Laws of Physics, Horizontal Velocity is unaffected by Vertical Velocity. Thus you need to first find how long it will take the stone to hit the ground.

Assume:

Acceleration due to gravity = 9.8 m/s^2

Velocity due to gravity = 9.8x m/s, where x is seconds

Displacement due to gravity = 9.8x^2 m, where x is seconds

Set Displacement = to 78.4m

78.4m = 9.8x^2 m

and solve for..

x = sqrt(8) = 2 sqrt(2)

Then to find displacement horizontally, multiply time*velocity horizontally

distance from base of cliff = (2 sqrt(2))s * 8 m/s = 16 sqrt(2) m

Q: A stone is thrown horizontally at 8 ms from a cliff 78.4 m high how far from the base of the cliff will the stone strike the ground?

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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).

64 METERSA+

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.

10 m/s

Answer: 44 meters

Related questions

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).

64 METERSA+

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.

10 m/s

Answer: 44 meters

Answer: 3 seconds

"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

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

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

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.