A student stands at the edge of a cliff 50 m above a beach and throws a stone horizontally over the edge with a speed of 18 m/s. How long after being released does the stone strike the beach below the cliff (take g = 10 m/s2)
Answers
this is very easy problem
we have given
distance=s=50m
and speed=v=18m/s
we have to find
t=?
we will use formula
s=vt
t=s/v
t=50m/18m/s
t=2.78 seconds
i want to teach another another trick here
we can see that meter in quotient and denominator will cancel out each other and seconds and in double denominator so that will become quotient hence we can check from that unit of time is second.
i hope you got it
if you have any confusion feel free to contact me
The time taken by stone to strike the beach is 3.16 seconds.
Calculation of Time of Flight for a Projectile Launched Horizontally from a Cliff
Displacement in the vertical direction is given by
=> y = ut + 1/2 gt²
Where
y is the displacement = 50 m
u is the initial vertical velocity = zero in this case ]
g is the acceleration due to gravity = 10 m/s²
t is the time taken
Substitute the above values in the equation we get
=> 50 = 0 + 1/2 × 10 × t²
=> 50 = 5t²
=> t² = 10
=> t = √10
=> t = 3.16 sec
Therefore
The time taken by stone to strike the beach is 3.16 seconds.
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