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ICSE Class 9 Answered

 A ball is thrown with the speed of 49 M per second upwards calculate the time after which it returns to the thrower. Solve for g=10m/s^2
Asked by bsap6777 | 29 Sep, 2018, 03:57: PM
answered-by-expert Expert Answer
A  ball thrown upward with velocity 
u= 49m/s
When ball reaches highest point v = 0
 
by first law of equation of motion,
v= u +gt 
g = -10 m/s2 (ball goes against gravity)
 
v= 49 + (-10) x t
t = 49/10 = 4.9 s
 
Considering, time of ascent = time of descent
Thus, total time required for ball to reach ground is 4.9 + 4.9 = 9.8 seconds
Answered by Shiwani Sawant | 01 Oct, 2018, 12:56: PM

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