Problem 12: A certain juggler usually tosses balls vertically to a height 3.8 yd. To what height must they be tossed if they are to spend twice as much time in the air?
Problem 12: A certain juggler usually tosses balls vertically to a height 3.8 yd. To what height must they be tossed if they are to spend twice as much time in the air?
Solution:
Given for the toss is;
Vertical distance=3,8 yards=3.8×0.9144= 3.4747meters
We know gravitational acceleration = g= 9.8 meter/ sec2
We find time for this distance using;
Where
S= distance=3.4747 m
Vi = initial velocity =0 (because ball is starting from rest position)
a=g=gravitational acceleration= 9.8 m/sec2
Putting these values in abov relation;
Now if we want it go stay double the time in air- we use;
this time with;
Vi=0 m/sec
S= distance
t=0.84209408×2=1.684188 sec (we multiply time by 2 to get double duration)
a=g=9.8 m/sec2
Putting values in relation;
So, it should be tossed up 15.2 yards high to remain double the time in air