Asked by Lee
John stand at the top of a 30m building. He throws a stone upwards with a speed of 5m/s. Calculate. (1)how long the stone is in the air. (2)the maxi
mum height the stone reaches the ground?
mum height the stone reaches the ground?
Answers
Answered by
Anonymous
Start by solving for the height the stone reaches at the top of its trajectory. You can use the kinematic equation Vf^2=Vi^2+2gd where g is the acceleration due to gravity (I used 9.8) and d is the displacement. Knowing that the velocity in the y direction is 0 m/s when the rock is at the max y displacement, you can solve for d. Don't forget that g is negative. This will be the height the rock travels ABOVE the building. It comes out to be something like 1.28 m, so add 30 m and you have part 2.
For part two you should use the kinematic that is d=ViT+0.5GT^2. You'll have to use the quadratic equation to solve for t, but that's OK. Use the answer to part two (add the 30 m) for d and the 5 m/s for Vi. The positive solution is your answer. I got about 2 seconds.
Hope this helps!
For part two you should use the kinematic that is d=ViT+0.5GT^2. You'll have to use the quadratic equation to solve for t, but that's OK. Use the answer to part two (add the 30 m) for d and the 5 m/s for Vi. The positive solution is your answer. I got about 2 seconds.
Hope this helps!