if the distance travelled by a freely falling body in the last second of its journey is equal to distance travelled in the first 2s, the time of descent of the body is
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157
Distance travelled in 1st 2 seconds
S = 0.5at^2
= 0.5g (2^2)
= 2g
Distance travelled in last second
S = g(n - 1/2)
Is it given that S = 2g
2g = g(n - 1/2)
n = 2 + 1/2
n = 2.5 second
Time of fall is 2.5 seconds
S = 0.5at^2
= 0.5g (2^2)
= 2g
Distance travelled in last second
S = g(n - 1/2)
Is it given that S = 2g
2g = g(n - 1/2)
n = 2 + 1/2
n = 2.5 second
Time of fall is 2.5 seconds
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7
Answer:
2.5 second
Explanation:
फाइंड डिस्टेंस ट्रैवल्ड इज इन फर्स्ट 2 सेकंड ग्रेड फाइंड डिस्टेंस ट्रैवल्ड इन द फर्स्ट टू सेकंड एंड डिस्टेंस ट्रैवल्ड इन लास्ट सेकंड दिन वी कैन गेट द आंसर
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