d 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? | Job Exam Rare Mathematics (JEMrare) - Solved MCQs

# 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