Given,
The initial velocity of the projectile, v₀=2.5 m/s
The height of the table, y₀=0.8 m
As the marble was shot horizontally, the angle of the projection will be zero, i.e., θ=0°
a)The x-component of the initial velocity is given by,
[tex]v_{0x}=v_0\cos \theta[/tex]On substituting the known values,
[tex]\begin{gathered} v_{0x}=2.5\times\cos 0^{\circ} \\ =2.5\text{ m/s} \end{gathered}[/tex]Thus the x-component of the velocity is 2.5 m/s
b) The y-component of the initial velocity is given by
[tex]v_{0y}=v_0\sin \theta[/tex]On substituting the known values,
[tex]\begin{gathered} v_{0y}=2.5\times\sin 0^{\circ} \\ =0\text{ m/s} \end{gathered}[/tex]Thus the y-component of the velocity id 0 m/s
c) As there is no force acting in the x-direction the acceleration in that direction will be zero.
d) In the y-direction the gravity is acting on the marble throughout its flight. Thus the acceleration of the marble in y-direction will be equal to the acceleration due to gravity.
Thus the magnitude of the acceleration in the y-direction will be equal to 9.8 m/s²
e) Applying the equation of the motion in the y-direction,
[tex]y=y_0+v_{0y}t+\frac{1}{2}gt^2[/tex]Where y is the final height of the marble, which is zero meters.
As we have taken height as positive, that is positive y-direction is upwards, the acceleration due to gravity will be negative, as its direction is downwards.
On substituting the known values in the above equation,
[tex]\begin{gathered} 0=0.8+0+\frac{1}{2}\times-9.8\times t^2 \\ \Rightarrow t^2=\frac{-0.8\times2}{-9.8} \\ \Rightarrow t^2=0.16 \\ \Rightarrow t=0.40\text{ s} \end{gathered}[/tex]Therefore the time of flight is 0.40 s
f) The displacement of the marble in the x-direction, i.e., the length traveled by the marble along the floor is given by,
[tex]x=v_{0x}t[/tex]On substituting the values,
[tex]\begin{gathered} x=2.5\times0.4 \\ =1\text{ m} \end{gathered}[/tex]Therefore the distance traveled by the marble is 1 m
