An Object is projected upwards from a building of height 100m with a velocity of 5m/s. Find (i)time taken by the object to reach the ground (ii) velocity of the object while striking the ground
Answers
Answer:
The time taken by the object is approximately 10s +10s = 20 seconds
The time taken by the object to reach the ground , from the highest point, t2 = √( 2h/g)= √(2*101.27/9.81)=4.5438 s = 4.544 s. Total time to reach the ground= t1 +t2= 0.51 + 4.54 = 5.05 s.
Given: height 100m with a velocity of 5m/s
To find: Time and velocity.
Solution:
Displacement is outlined as the change available of an object. It is a heading size and has a management and size. It is presented as an missile that points from the offset position to the ending position. For example- If an object moves from A position to B, before the object’s position changes. This change available of an object is famous as Displacement.
Distance and dislocation are two quantities that appear to mean the alike but are clearly various accompanying various significations and definitions. Distance is the measure of “by what method much ground an object has enclosed all the while allure motion” while dislocation refers to the measure of “by virtue of what unusual of place is an object.” In this item, allow us believe the distinctness 'tween distance and dislocation.
Let simplify it,
V²—U²= 2AS
0–5²= —2×10S
S= 25/20= 5/4 M= 1.25 M
V= U+AT
U= 0
A= G= 10
S= 100+1.25= 101.25
S= UT+(1/2)AT²
101.25= (1/2)×10T²
T²= 101.25/5= 20.25
T=4.5 S
Total time to reach ground= .5+4.5= 5 S
V= U+AT= 0+10×4.5= 45 M/S
Time taken to reach ground= 5S
Velocity at ground= 45M/S
Hence we get v = 45 and t as 5s.
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