A ball is projected vertically up from ground with a speed of $40 {ms}^{- 1}$ $(g = 10 {ms}^{- 2})$

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Maximum height is $80 m$

At time $t = 5 s$ , it is at a height of $75 m$ from the ground

From $t = 3 s to t = 6 s$ , its displacement is $10 m$ downwards

While moving up, velocity vector points upwards and acceleration vector points downwards; while moving down, velocity vector points downwards and acceleration vector points downwards

answer is A, B, D.

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Detailed Solution

Maximum height, $H = \frac{u^{2}}{2 g} = \frac{40^{2}}{2 \times 10} = 80 m$ . Option (a) is correct.
Taking origin on ground and upward vectors as positive
At time $t = 5 s$ , $h = u t - \frac{1}{2} g t^{2} = 40 (5) - \frac{1}{2} (10) {(5)}^{2} = 200 - 125 = 75 m$ Option (b) is correct.
$h_{3} = 40 (3) - \frac{1}{2} (10) {(3)}^{2} = 120 - 45 = 75 m$

$h_{6} = 40 (6) - \frac{1}{2} (10) {(6)}^{2} = 240 - 180 = 60 m$
From $t = 3 s to t = 6 s$ , its displacement is $h_{6} - h_{3} = 60 m - 75 m = - 15 m$ Option (c) is wrong.
Acceleration due to gravity always points downwards.
Velocity vector points in the direction of motion. Option (d) is correct.