Question

Solve question 12
Please if u can

Answer 1

(a)-

use,
$s = u \times t - \frac{1}{2} \times \ g \times {t}^{2}$
s is the total displacement of the ball=0
u is the initial velocity.
t is the time
g is the acceleration due to gravity= 9.8 m/sec square

put the value

0=u×4 - 1/2×9.8×4×4
u×4= 9.8×2×4
u×4 = 78.4
u = 78.4/4
u = 19.6 m/sec

(b)-

At maximum height velocity of the ball=0, v=0

use, v^2=u^2-2×g×s
0 = 19.6×19.6 - 2×9.8×s
19.6×19.6 = 19.6×s
s= 19.6 m


(c)-

use,
s=u×t-1/2×g×t^2

s= 19.6×3 - 1/2×9.8×3×3

s= 58.8 - 4.9×9
s= 58.8 - 44.1

s= 14.7 m above the ground..


hope u understand

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