N apple drops from a tree and hits the ground in 1.5 seconds. what is its speed just before it hits the ground?
Working formula is Vf - Vo = gT where Vf = velocity of apple just before hitting the ground Vo = initial velocity of apple = 0 T = time for apple to hit the ground = 1.5 sec (given) g = acceleration due to gravity = 9.8 m/sec^2 (constant) Substituting appropriate values, Vf - 0 = 9.8(1.5) Vf = 14.7 m/sec. Hope this helps.
An apple falls from a tree and hits the ground in one second. What is its speed when its hits the ground?
a = -g = -9.8 m/s²; acceleration of gravity v₀ = 0 m/s; initial velocity t = 1 s; time v₁ = ? m/s; finial velocity y₀ = ? m; initial height y₁ = 0 m; finial height is on the ground ➊v₁ = v₀ + at ; use this equation v₁ = 0 + -9.8m/s²(1s) = -9.8 m/s ➋ y₁-y₀ = v₀t + ½at²; use this equation to find out how high 0 - y₀ = 0 - ½(9.8m/s²)(1s)² y₀ = 4.9 m high ⊽ = (y₀-y₁)/(t₀ - t₁); use this equation for average speed ⊽ = (4.9 m - 0 m)/(0s - 1s) = -4.9 m/s, average GENERAL EQUATIONS FOR MOTION WITH CONSTANT ACCELERATION ⊽ = (x₀ - x₁)/(t₀ - t₁) Average speed ➊v₁ = v₀ + at; use this when you don't know (y₁-y₀) finial and initial position, or at least their difference. ➋y₁-y₀ = v₀t + ½at²; use this when you don't know (v₁) finial velocity, at a given final point y₁. ➌v₁² = v₀² + 2a(y₁-y₀); use this when you don't know (t) time of travel between the two points (y₁-y₀). ➍y₁-y₀ = ½(v₀ + v₁)t; use this when you don't know (a) acceleration. ➎y₁-y₀ = v₁t - ½at²; use this when you don't know (v₀) initial velocity.
An apple falls from a tree and hits the ground 10 m below. With what speed will it hit the ground? The acceleration of gravity is 9.8 m/s2?
Potential at 10 m high = Kinetic energy on impact mgh = 1/2mv^2 v = √2gh = √ (2 x 9.8 x 10) = 14 m/s
An apple falls from a tree and hits the ground 12.4 m below?
with what speed will it hit the ground? The acceleration of gravity is 9.8 m/s^2. answered in utins of m/s I've tried a 100 times and I just can't get it :/
Physics, An apple falls from a tree and hits the ground 5 meters below. It hits the ground with a speed of...?
Hello you can set potential energy equal to kinetic energy potential energy of the apple when it still on the tree Epoz = mgh kinetic energy before it hits the ground: Ekin = 1/2 mv^2 Epot = Ekin mgh = 1/2 mv^2 gh = 1/2 v^2 v^2 = 2gh v = √(2gh) v = √(2*9.81*5) v = 9.9 m/s <--- ans. Regards
An apple falls from a tree and hits the ground 5 meters below. It hits the ground with a speed of?
I used d=1/2 gt^2 with d=5m and g=10 m/s^2 and obtain t=1. Since I used d=5, doesn't this mean the object traveled a distance of 5 m in one second and thus its speed is 5m/s. (The correct answer is somehow 10 m/s)
What forces act on an apple as it falls down from a tree?
There is two forces acting on that Apple .The Gravitational Force which is the reason why the apple falling towards the center of earth.And the second less powerful force called Air drag or friction which try to keep the apple stable but Because of the power of Gravitational force and the air is not that dens so it just dont affect the apple that much as the first force does so as usual apple fall down.