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Grade 11General Physics

A ball is thrown vertically upward from the top of the tower with a speed of 40m/s returns back to the ground level in 10s .The height of the tower is

Profile image of Rahul
9 Years agoGrade 11
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1 Answer

Profile image of PASUPULETI   GURU MAHESH
9 Years ago
Top of the tower u=40 
v=u-gt
t=40/10=4 sec
maximum height h= ut -1/2 gt2
 you get height ball reach from the tower.
and calculate the height from the maximum height
      H=ut+1/2gt2
initial at maximum height u=0
then H =1/2gt2
then the tower height = H-h
you calculate and get the height of the tower
 
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