An object is thrown straight up with a speed of 30m/s. How high above the point of release is it and what is its velocity after 2s and 6s? How high does it rises? (Please send me the solution with an attachment)
Answers
Given u=30 m/s v=0 g=10 m/ss
At t=2 s
v=u-at (since it is thrown upwards)
=30-10*2
=30-20
=10 m/s
S=ut-1/2*a*t*t
=30*2-5*2*2
=60-20
=40 m
At t=6 s
v=u-at
=30-10*6
=30-60
=-30 m/s (since velocity is negative so it means it is coming in opp.
direction)
S=ut-1/2*a*t*t
=30*6-5*6*6
=180-180
=0 m (since displacement is 0 it means the body has covered a certain
distance, twice)
So the time to reach max. distance = 6/2 = 3 s
max. height covered = u*3-1/2*a*3*3
= 30*3-5*9
= 45 m
This is a basic kinematics question, where acceleration is constant, so SUVAT can be used.
s = ?
u = 20 m/s
v = 40 m/s
a = 9.81 m/s² (acceleration due to freefall)
t = ?
The most appropriate equation here is v² = u² + 2as
So 40² = 20² +2*9.81*s
→ 1600 = 400 + 2*9.81*s
→ 1200 = 2*9.81*s
→ 600 = 9.81*s
→ s = 61.1620795
s = 61.2 m (to 3 sf)
The building is 61.2 m high.
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