Aball is thrown horizontally at a height of 2.2 meters at a velocity of 65m/s off a cliff. assume no air resistance. how long until the ball reaches the ground?

We have that A ball is thrown horizontally at a height of 2.2 meters at a velocity of 65m/s off a cliff. Assuming no air resistance will take t to reach the ground

t=0.670sec

From the question we are told that

A ball is thrown horizontally at a height of 2.2 meters

a velocity of 65m/s off a cliff

Generally the equation for the Time is mathematically given as

s=ut+1/2gt^2

Therefore

Therefore

A ball is thrown horizontally at a height of 2.2 meters at a velocity of 65m/s off a cliff. Assuming no air resistance will take t to reach the ground

t=0.670sec

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The horizontal motion has no effect on the vertical drop.

From a drop, the distance the ball falls in 'T' seconds is

D = 4.9 T^2

so

2.2 = 4.9 T^2

T^2 = 2.2/4.9

T^2 = 0.449 sec^2

T = 0.67 second

i would like to make dinner for my mom, dad, and sister

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg