in ordrr to throw a ball staight up you push it with your hand from 0.5m above t

Question

in ordrr to throw a ball staight up you push it with your hand from 0.5m above the ground to 1.8m above the ground with acceleration 4.5 . How fast is the ball moving at the end of the push What is the maximum height reached by the ballAt what speed will the ball hit the ground in ordrr to throw a ball staight up you push it with your hand from 0.5m above the ground to 1.8m above the ground with acceleration 4.5 . How fast is the ball moving at the end of the push What is the maximum height reached by the ballAt what speed will the ball hit the ground

Explanation / Answer

initial speed=0

acceleration=4.5 m/s^2

distance travelled during this acceleration phase=1.8-0.5=1.3 m

using the formula,final speed^2-initial speed^2=2*acceleration*distance

==>final speed^2-0^2=2*4.5*1.3

==>final speed=sqrt(2*4.5*1.3)=3.4205 m/s

so the ball will be launched at 1.8 m height from ground at 3.4205 m/s

after it is launched, it will have a vertical downward accleration of 9.8 m/s^2 till it reaches maximum where its speed will be 0.

so if time taken for the speed of the ball to be 0 is t seconds,

then 0=initial veloicty+acceleration*t

==>0=3.4205-9.8*t

==>t=0.349 seconds

during this time, height reached=initial speed*time+0.5*acceleration*time^2

=3.4205*t-0.5*9.8*t^2

=3.4205*0.349-0.5*9.8*0.349^2=0.59639 m

hence total maximum height reached=1.8+0.59639=2.39639 m

when the ball starts dropping from the maximum height, its initial speed=0

acceleration=9.8 m/s^2

distance to be covered=2.39639 m

then final speed when the ball reaches ground=sqrt(2*9.8*2.39639)=6.8534 m/s

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