solve if u r good at physics 11 and 12
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Ques11: Sol. Initially the ball is going upward u = –7 m/s, s = 60 m, a = g = 10 m/s2 s = ut+ ½ at^2
⇒ 60 = -7t 0+ ½ 10t^2
⇒ 5t^2 – 7t – 60 = 0
t = (7+ √(49-4.5 (-60) ))/(2 x 5) = ( 7 ± 35.34 )/10 (taking positive sign)
t = (7+ 35.34)/10 = 4.2 sec (∴ t ≠ -ve)
Therefore, the ball will take 4.2 sec to reach the ground.
Ques12:h =19.6m
u=0 g=9.8m/s
a) time taken to travel first meter
S=1m u=0
S=ut+1/2at^2
1=0t+1/2×9.8×t²
t=√1/4.9=0.45sec.
b) time taken to travel last meter
First we have to find velocity of body when it is 1m above the ground
V²-U²=2aS
U=0, S=19.6-1=18.6, g=9.8m/s
V²=2×9.8×18.6=19.1m/s
S=ut+1/2at²
1=19.1t+4.9t²
4.9t²+19.1t-1
On solving this quadratic equation we will get t=0.05sec.
⇒ 60 = -7t 0+ ½ 10t^2
⇒ 5t^2 – 7t – 60 = 0
t = (7+ √(49-4.5 (-60) ))/(2 x 5) = ( 7 ± 35.34 )/10 (taking positive sign)
t = (7+ 35.34)/10 = 4.2 sec (∴ t ≠ -ve)
Therefore, the ball will take 4.2 sec to reach the ground.
Ques12:h =19.6m
u=0 g=9.8m/s
a) time taken to travel first meter
S=1m u=0
S=ut+1/2at^2
1=0t+1/2×9.8×t²
t=√1/4.9=0.45sec.
b) time taken to travel last meter
First we have to find velocity of body when it is 1m above the ground
V²-U²=2aS
U=0, S=19.6-1=18.6, g=9.8m/s
V²=2×9.8×18.6=19.1m/s
S=ut+1/2at²
1=19.1t+4.9t²
4.9t²+19.1t-1
On solving this quadratic equation we will get t=0.05sec.
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