Ok so I have to use kinematic equations to solve these. But I make like a X and Y chart Ex...X|Vix=2.89 dx=? t=? Y|Viy=0 dy=3.7 and so forth.
1. A cliff diver strikes the water 10.3 meters from a cliff 25 meteres high. At what horizontal speed did he leave the cliff?
2. A ball rolls off a level table that is 3.7m high at 2.89 m/s. How far away from the table will it stike the floor?
3. A car leaves a cliff at 55 mph( this part needs converting) and strikes the flat ground below 20m from the cliff. How high is the cliff?
4.A rock is thrown straight up at 18m/s from the top of a 31m high building, goes straight up and then straight down to the street below. Using only one kinematic equation, find the time it takes to reach the street.
1. 300 mph
2. 41.675 cm
3. 759.3547M high
4.Vix=2.89 dx=4.9 Viy=18.315
hope this helps!
im only in yr 8 at school by the way
To solve these problems, you can use the kinematic equations, which are a set of equations that describe the motion of an object under constant acceleration.
1. For the first problem, you can use the equation: dx = Vix * t. You are given dx = 10.3m and dy = 25m. Since the motion is two-dimensional (both horizontal and vertical), you can separate it into two independent one-dimensional problems. Plug in the values you know into the equation to solve for Vix, the initial velocity in the x-direction.
2. For the second problem, you can use the equation: dy = Viy * t + (1/2) * a * t^2. You are given dy = 3.7m, Viy = 2.89m/s, and a = 9.8m/s^2 (acceleration due to gravity). Set dy equal to zero, as the ball will strike the floor at that point. Solve for t, the time it takes for the ball to hit the floor. Then, use the equation dx = Vix * t to find the horizontal distance it will strike from the table.
3. For the third problem, you need to convert the speed from mph to m/s. To do this, remember that 1 mph is equal to 0.44704 m/s. Once you have the speed in m/s, you can use the same method as in problem 1 to find the height of the cliff. Use the equation dx = Vix * t, where dx = 20m and Viy = 0 since the car does not move vertically.
4. For the fourth problem, since the rock goes straight up and then straight down, the total time it takes to reach the street is twice the time it takes to reach the top of the trajectory. Use the equation: Vf = Vi + a * t, where Vf = 0 at the top of the trajectory. Solve for t and double the result to find the total time from top to bottom.