an object is thrown up with a speed of 50m/s. It's height from the ground after 7 is
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An object is thrown up with a speed of 50 m/s. What is its height from the ground after 7 s?
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1 Answer

Nitish Kumar, B.Tech CSE from National Institute of Technology, Hamirpur (2019)
Answered 4 years ago
Using 2nd equation of motion, S= ut+1/2gt^2, we get
S = 50*7–4.9(49) = 350–240.1=109.9m.
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Bhabindra Kunwar Bhavabhinna, PhD Scholar from Tribhuvan University (2017)
Answered 7 months ago
An object with an initial velocity of 12 m/s west experiences a constant acceleration of 4m/s2 west for 3 seconds. During this time, what distance does the object travel?
Assumption from the question:
Initial velocity (u) = 12 m/s
Acceleration (a) = 4 m/s^2
Praduryantar (r) = 4 m/s
Time (t) = 3 s
No. of ref. points (n) = 4
The distance traveled in 3 seconds (s) =?
Solution:
(a) Using suvat equations,
s = ut+at^2 = (12*3+(1/2)4*3*3) m = (36+18) m = 54 m
(b) Using suvrt equations,
s = ut+nrt/2 = (12*3+4*4*3/2) m = (36+24) m = 60 m
Test:
The object will travel 3 different distances in each 3 time interval respectively.
This means,
Distance traveled in first 1 second (say d) = 12+4 m = 16 m.
Distance traveled in second 1 second (say e) = 12+4+4 m = 20 m.
Distance traveled in third 1 second (say f) = 12+4+4+4 m = 24 m
Experimentally, the total distance traveled in 3 seconds (s) = d+e+f = 60 m
Conclusion:
Using SUVRT equations we will get (s) = 60 m
Using SUVAT equations we will get (s) = 54 m
From experiment, we will get (s) = 60 m
Here are two different results for the same event. This is a kind of interesting beauty of our physics.
I hope, this answer will also be helpful like others. Thank you.