From the top of a cliff 50m above sea level, a stone is thrown vertically upwards with a velocity of 15m/s. After how many seconds and with what velocity will it hit the sea
Answers
Given : From the top of a cliff 50m above sea level, a stone is thrown vertically upwards with a velocity of 15m/s.
To Find : After how many seconds and with what velocity will it hit the sea
Solution:
S = ut + (1/2)gt²
S = 50 m
u = -15 m
g = 10 m/s²
50 = -15t + (1/2)*10t²
=> 5t² - 15t - 50 = 0
=> t² - 3t - 10 = 0
=> (t - 5)(t + 2) = 0
=> t = 5 as time can not be negative
Hence after 5 secs it hit the sea
v = u + at
v = -15 + 10 * 5
v= 35 m/s
It will git with a velocity of 35 m/s
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SOLUTION
GIVEN
From the top of a cliff 50m above sea level, a stone is thrown vertically upwards with a velocity of 15m/s.
TO DETERMINE
After how many seconds and with what velocity will it hit the sea
EVALUATION
Here by the given -
s = 50 m
u = - 15 m/s
a = 10 m/s²
t = Time
We know that
v² = u² + 2as
⇒v² = 225 + 2 × 10 × 50
⇒v² = 225 + 1000
⇒v² = 1225
⇒ v = 35
The required velocity = 35 m/s
Again we have
v = u + at
⇒ 35 = - 15 + 10t
⇒ 10t = 50
⇒ t = 5
Hence the required time = 5 sec
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