fungulo Posted April 16, 2008 Posted April 16, 2008 How would i Figure out this question? A skydiver steps out of a helicopter hovering thousands of metres above ground. After 3.5 s of free fall the skydiver pulls the rip cord. What is the skydivers displacement during the 3.5 s of free fall? Please tell me step by step because i have tried everything to get this awnser and i cant get it thank you in advance!!
Cap'n Refsmmat Posted April 16, 2008 Posted April 16, 2008 Are you familiar with the equation [math]x = \frac{1}{2}at^2[/math] where x = displacement, a = acceleration, and t = time elapsed? The version you've seen might look different, but it should at least resemble this. You should know a (it's gravitational acceleration, which is the same everywhere), so you can just plug in the numbers and get an answer.
Cap'n Refsmmat Posted April 16, 2008 Posted April 16, 2008 What's the standard acceleration due to gravity? (Hint: It's always 9.8 m/s2)
fungulo Posted April 16, 2008 Author Posted April 16, 2008 Thanks!! Another Question!! Based on a reaction time of 1.50 s and a braking rate of - 5.85 m/s the stopping distance of a vehicle initially travelling 90 km/h would be? Step by Step please!!
Cap'n Refsmmat Posted April 16, 2008 Posted April 16, 2008 First, you'd want to convert the vehicle's velocity to meters per second. Next, you want to see how far the car traveled before the driver hit the brakes. The equation for that is something like this: [math]x = vt[/math] The distance = velocity * time. The distance traveled after the driver hits the brakes is weirder. The equation I've used is this: [math]V^2 = V_0^2 - 2a(x - x_0)[/math] Where V2 = the final velocity (squared) (which is 0, because the car is stopping), V0 = the initial velocity of the car, a = the acceleration of the car (the breaking rate), and x - x0 is the distance it travels.
fungulo Posted April 16, 2008 Author Posted April 16, 2008 Thanks i got the awnser!!! A high performance car initially travelling 97.2 km/h came to a stop in just 2.9 s. The mass of the car and its contents was 1850 kg. what was the magnitude of the average net force on the car while it decelerated? Step by step please!!!
Cap'n Refsmmat Posted April 16, 2008 Posted April 16, 2008 Find the acceleration the car is experiencing. You'll need an equation that relates velocity and acceleration: [math]v = at + v_0[/math] where v = final velocity, a = acceleration, t = time, and v0 = initial velocity. (You should have an equation like this somewhere.) Just plug and chug. You should hopefully know that force = mass x acceleration, so you just stick in the numbers and get the answer.
fungulo Posted April 16, 2008 Author Posted April 16, 2008 Thanks so much!!! I have another one thow!! A ball with a mass of 1.50 kg travelling +2.00 m/s collides with a stationary ball with a mass of 1.00 kg. After the collision the velocity of the 1.50-kg ball is +0.40 m/s. what is the velocity of the 1.00-kg ball after the collision? THanks
Cap'n Refsmmat Posted April 16, 2008 Posted April 16, 2008 Think about conservation of momentum. The total momentum before has to be the same as the total momentum after the collision.
fungulo Posted April 16, 2008 Author Posted April 16, 2008 ok this is the final question i have!!! a 0.080-kg egg test dummy is fitted with a helmet and attached to a swing. this egg test dummy is pulled back and released allowing it to collide with a cement block. the impulse on the egg test dummy is -0.39 N.s over an interval of 0.050 s. what is the magnitude of the force on the egg test dummy during this time interval? Step by Step please. thanks.
Cap'n Refsmmat Posted April 16, 2008 Posted April 16, 2008 Look at the equation you use to find impulse. You already have the right pieces to use in it.
thedarkshade Posted April 16, 2008 Posted April 16, 2008 Look at the equation you use to find impulse. Care to share a link
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