Help Please?

Question

A ball is thrown upward from the roof of a building 100 m tall with an initial velocity of 20 m/s. Use this information for exercises 35 to 38. When will the ball reach a height of 80 m?

Answer 1

Ans: 2 sec
u have to take gravitation into account. Assume ur origin at the top of building and positive direction upwards. Use the formula: x=x0+u*t+0.5*g*t*t

in this case, x0=0 (as ur orgin is at top of building), x=-20(80 m above ground is -20 wrt ur origin), u=20 and g=-10

puttin these values u will get (t-2)(t-2)=0 => t=2 sec.

For better understandin of these concepts refer to HC verma part1

Hope this helps

Answer 2

using V2 = u2+ 2gh
U=20m/s g= -10 as the ball goes up. At the highest point v = 0
Therefore 400= 20h or h=20m
Now v= u- gt , v=o or h= ut - ½ gt2 ( solve quadratically)
Therefore time taken for 20m is 20/10 = 2 seconds
To reach 80 m above ground the ball has to travel 40 m from the highest point. Using the relation h= ut + ½ gt2 we see u= 0 as it is at top
Therefore h=40 g=10 so t2=8 so t= 2.83 seconds.
So total time taken is 4.83 seconds

Answer 3

100 / 20 = Amount of time to travel all the way up.
80 / 20 = Amount of time to travel 80 meters.

Answer 4

at 1 sec.
you can do this intuitively.
it starts at 100m tall, but it falls at a rate of 20m/s. So every second, it goes down 20m. so you reach the height of 80m in 1 sec.

Answer 5

5+1=6 sec.!