A diver running 1.7 m/s dives out horizontally from the edge of a vertical cliff and reaches the water?
A diver running 1.7 m/s dives out horizontally from the edge of a vertical cliff and reaches the water below 2.9 s later. How high was the cliff and how far from its base did the diver hit the water? ____m (cliff) ____m (distance from base)
A diver running 1.8 m/s dives out horizontally from the edge of a vertical cliff ?
A diver running 1.8 m/s dives out horizontally from the edge of a vertical cliff and 3.0 s later reaches the water below. How high was the cliff, and how far from its base did the diver hit the water?
A diver running 1.6 m/s dives out horizontally from the edge of a vertical cliff and 2.6 s later reaches the w?
His horizontal and vertical motion are independent. vertical: y = y0 + v0 t + 1/2 a t^2 (a = 9.8 m/s^2) His vertical starting speed is zero. Lets call the top of the cliff y = 0 then y = 0 + 0 - 1/2 9.8 2.6^2 = -33 m, so the cliff is 33 m high. His horizontal speed remains 1.6 m/s (ignoring friction) So he hits the water 2.6s * 1.6 m/s = 4.16 m out.
A diver running 1.0m/s dives out horizontally from the edge of a vertical cliff and 3.5 s later reaches the?
There are two parts to this question: 1) The vertical component, and 2) the the horizontal component. 1) For the vertical component, the question resolves itself into: How far do you fall if you fall freely from rest under gravity for 3.5.seconds ? Use s = ut + 1/2 a t^2, where s is the distance fallen, u the initial velocity, t the time taken, and a the acceleration due to gravity, so, u = 0, t = 3.5, and a = 9.8 m/sec/sec, so: s = 0xt + 1/2 x 9.8 x 3.5 x 3.5, so s = 0 + 4.9 x 12.25, so s = 0 + 60.025m. So the clifff is 60.025 m high. 2) The diver moves horizontally for 3.5 secs at a speed of 1.0m/sec, so the horizontal distance travelled is 3.5 x 1 = 3.5m. So the diver hit the water 3.5m from the base of the cliff. Hope this helps, Twiggy.
Homework Question: An 8.0 kg toy is dropped from a height of 7.0 m. What is the kinetic energy of the toy just before it hits the ground?
Well, our toy has an energy E, which is the sum of the potential and kinetic energy.So, when the toy is at the top, 7 m, it has potential energy, because it is at rest (its speed is 0)Ep = m*g*hWhere m is the mass (8 kg), g is the Earth's gravity (9.81 N/kg) and h is the height (7m). Please note that sometimes teachers use g = 10, in order to simplify the calculations.For the sake of simplicity, I will use 10. If you want to be exact, use 9.81.So, plugging in our numbers, we get:Ep = 8 kg * 7 m * 10 N/kg = 560 JWhen the toy falls, its energy is transforming from potential to kinetic.So, when the height is 0, all the energy is kinetic.Ek = 1/2 * m * v^2So, we know that energy is conserved, which means that the initial energy is the same as the final energy. We had calculated the initial energy to be 560 J, which means that Ek is also 560JSubstituting, we get that:1/2 * 8 kg * v^2 = 560V^2 = 140So, v is the square root of 140, which is 11.83 m/sFor any questions, feel free to comment!