Q A ball dropped from height 'h' reaches the ground in time T. What is its height from the ground at time T/2?

h/28

h/4

h/2

3h/4

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Solution

Dear Student
$44 . S = u t + 1 / 2 g t^{2} h = 1 / 2 \times g \times t^{2} h' = 1 / 2 \times g \times (\frac{t}{2})^{2} h' = 1 / 2 \times g \times t^{2} / 4 h' = h / 4 F r o m b o t t o m = h - h / 4 = 3 h / 4 O p t i o n 4 R e g a r d s$

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Q ​A ball dropped from height 'h' reaches the ground in time T. What is its height from the ground at time T/2? h/28 h/4 h/2 3h/4

Dear Student 44 . S=ut+1/2gt2h=1/2×g×t2h'=1/2×g×(t2)2h'=1/2×g×t2/4h'=h/4From bottom=h-h/4=3h/4Option 4Regards

Q A ball dropped from height 'h' reaches the ground in time T. What is its height from the ground at time T/2?

h/28

h/4

h/2

3h/4

Dear Student
$44 . S = u t + 1 / 2 g t^{2} h = 1 / 2 \times g \times t^{2} h' = 1 / 2 \times g \times (\frac{t}{2})^{2} h' = 1 / 2 \times g \times t^{2} / 4 h' = h / 4 F r o m b o t t o m = h - h / 4 = 3 h / 4 O p t i o n 4 R e g a r d s$