Answer to Question #202173 in Mechanics | Relativity for BUTCH MANRIQUE

Question

A marble is thrown horizontally from a tabletop with a velocity of 1.50 m/s. The marble falls 0.70 m away from the table’s edge. How high is the lab table? What is the marble’s velocity just before it hits the floor? 

Answer 1

Gives

Velocity (v_0x)=1.50m/sec

d=0.70m

We know that

$d=V_{0x}t$

$0.70=1.50\times t$

t=0.47 sec

$V_{fy}=V_{0y}+gt$

$V_{fy}=0+(-9.8)\times0.47$

$V_{fy}=-4.5783m/sec$

$V=\sqrt{V_{ox}^2+V_{fy}^2}$

Put value

$V=\sqrt{{1.50}^2+{(-0.47)}^2}$

V=4.82m/sec