Question

an object is thrown up with a speed of 52m/s.its height from the ground after 7s is

Answer 1

H(s) = ut + 1/2at^2 

here , s = h , a = -g and t = 7sec 

so H = ut - 1/2gt^2
     H = 52 × 7 - 1/2 × 10 × (7)^2
     H = 364 - 5 x 49
     H = 364 - 245
     H = 119 m

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Answer 2

by second equation of motion

=> S = ut + ½at²

=> S = 52×7 + ½×(-10) ×49

=> S = 364 – 5×49

=> S = 364 – 245

=> S = 119 m

object height from the ground is 119 meter.
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hope it will help you .....✌✌
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