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
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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;

S=Vit+1/2a t^{2}


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;

3.4747=0\times t+\frac{1}{2}\times9.8 \times t ^{2}

\Rightarrow 3.4747=4.9 \times t ^{2}

\Rightarrow t ^{2}=0.70912245

\Rightarrow t =0.84209408 sec

Now if we want it go stay double the time in air- we use;

S=Vit+1/2a t^{2}

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;

S= 0\times(1.684188)+\frac{9.8}{2} \times (1.684188) ^{2}

S= \frac{9.8}{2} \times 2.8365

S= 13.9 meters=15.2 yards

So, it should be tossed up 15.2 yards high to remain double the time in air

 

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