Question

a boy throws a ball horizontally from a point 4.9m above horizontal ground.Determine the minimum speed at which the ball must be thrown to clear the fence 2.4m high at a horizontal distance of 8m from the point of projection. find also the distance beyond the fence of the point at which the ball strikes the ground if projected at this minimum speed.

Answer 1

It's given the boy is standing at 4.9 m above height, he throws it horizontally, we want the ball to clear the fence 2.4 m high at a horizontal distance of 8 meters.
Since the boy is only giving a horizontal velocity to the ball, ball is acted by gravity and has zero initial velocity in the vertical direction
Let's say after x second ball just crosses the fence
we know
\[S=ut+\frac{1}{2} at^{2}\]

s= 2.5 meters downward (4.9-2.4=2.5)
a=g=9.8 m/s^2
u=0
t=x
\[2.5= \frac{1}{2} 9.8* x^2 \]
we get
\[x^2 = \frac {5}{9.8} \]
we get
x=0.71 seconds
In 0.71 seconds the ball should cover 8 meters horizontally
so
horizontal velocity \(v\)
\[v=\frac{8}{0.71}= 11.267 \]
so the minimum horizontal velocity =11.267 m/s

Answer 2

let's find the time taken by ball to reach ground
s=4.9( in downward direction)
a=9.8m/s^2
u=0
\[s=ut+\frac{1}{2} at^2 \]
\[4.9= \frac{1}{2} 9.8 t^2 \]
t=1 second
in 1 second the horizontal velocity covered by ball
d=11.267*1= 11.267 meters
So the ball covers 11.267 meters horizontally before striking ground.

Answer 3

linardo did you understand?

Answer 4

can u draw diagram to me.. i think it's more clear to me to understand it..