Question

Determine the initial velocity of a marble that lands 1.8m away from a tabletop that is 0.8m from the ground.
Acceleration due to gravity = 9.85
0 2.25 m/s
2.6 m/s
0 1.8 m/s
4.5 m/s

Answer 1

Answer:

4.5 m/s

Explanation:

$s_x$ = Horizontal displacement = 1.8 m

$s_y$ = Vertical displacement = 0.8 m

$u_x$ = Horizontal velocity

g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

t = Time taken to fall down

$s_y=u_yt+\dfrac{1}{2}gt^2\\\Rightarrow 0.8=0+\dfrac{1}{2}\times 9.8\times t^2\\\Rightarrow t=\sqrt{\dfrac{2\times 0.8}{9.8}}\\\Rightarrow t=0.4\ \text{s}$

$s_x=u_xt+\dfrac{1}{2}a_xt^2\\\Rightarrow 1.8=u_x\times 0.4+0\\\Rightarrow u_x=\dfrac{1.8}{0.4}\\\Rightarrow u_x=4.5\ \text{m/s}$

The initial velocity of the marble is 4.5 m/s.