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A block slides off a horizontal table 4 meters high with a velocity of 12-m/s. Find: a) The horizontal distance from the table at which the...

      

A block slides off a horizontal table 4 meters high with a velocity of 12-m/s. Find:
a) The horizontal distance from the table at which the block hits the floor.
b) The horizontal and vertical components of the velocity when it reaches the floor.

  

Answers


Wilfred
(a) Initial vertical velocity = 0, a=10, S=4m
Therefore: S= ut + 1/2 gt2
4 = 0+5t2
t = 0.8944
Range (horizontal dist) = initial vel. (constant) x time
= 12 x 0.8944
= 10.73m

b) Initial and final horizontal velocities are equal =12m/s
Final vertical velocity = V= u + at
V= 0+10x0.8944
V= 8.944m/s
Wilfykil answered the question on March 26, 2019 at 09:04


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