A ball is thrown from the top of a building
of height 25m with an initial velocity of
15m/sec. If the height of the ball h from
ground at any point of time t is given by
h 25 + 10t - 3t. The time taken by the
ball to reach the ground is
Answers
Given : the height of the ball h from ground at any point of time t is given by h 25 + 10t - 3t².
To find : The time taken by the ball to reach the ground
Solution:
the height of the ball h from ground at any point of time t is given by
h = 25 + 10t - 3t²
at t = 0
h = 25 m
ball reach the ground => h = 0
=> 25 + 10t - 3t² = 0
=> 25 + 15t - 5t - 3t² = 0
=> 5(5 + 3t) - t(5 + 3t) = 0
=> (5 - t) (5 + 3t) = 0
=> t = 5 , t = -3/5
Hence after 5 secs ball will reach the ground
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Answer:
5/3
Step-by-step explanation:
h=25m,u=15m
h=25+10t-3t^2
dh/dt=0+10(1) -3(2t) =10-6t
therefore, dh/dt=0
10-6t=0
6t=10
t=5/3sec
