Physics scenarios Sarah is demonstrating her jumping abilities to a group of PHY
ID: 1328685 • Letter: P
Question
Physics scenarios
Sarah is demonstrating her jumping abilities to a group of PHYS 2210 students. For her first stunt (stunt #1), she steps off the top of a building from rest and lands on a cushion directly below. It takes a time At for her to hit the ground. For her second stunt (stunt #2), Sarah goes back to the top of the same building and takes a running leap off of it. Her initial velocity is in the horizontal direction (i.e. there is no vertical component to her leap). She lands on some cushions a horizontal distance D from the base of the building, as shown in the figure Stunt #1 Stunt #2 Scenario 4: Imagine a tower built so high that its top extends above the earth's atmosphere where air resistance (drag) can be neglected. The top of the tower is at a height H above the surface of the earth. A projectile of mass mp is fired off the top of the tower with speed vo. The initial direction of the projectile is tangential to a circular orbit, as shown in the figure 7 0 G = 6.67×10-11 m3/kg/s2 Mearth = 5.97×1024 kg G-Mearth = 3.98×1014 m3/s2 Rearth = 6.37×106 m earthExplanation / Answer
There are multiple questions here . i am allowed to answer only 1 at a time. I will answer question 1 for you.Please ask other as different question.
In stunt #1 as well as stunt#2, vertical component if initial vecolity was 0 in both cases. So time taken to hit ground will be same in both cases which will de delta t.
To calculate V, use Newton's law in horizontal direction.
V = horizontal distance / time
= D/delta t
If you need to calculate final velocties and angle thetha.
Horizontal component of final velocity will not change. I will still be D/delta t
Vertical coponent will increase to g*delta t
so net velocity will be sqrt((g*delta t)^2 + (D/delta t)^2)
angle thetha = atan (d*delta t / (D/deta t))
atan (d*(delta t)^2/ D)