Respuesta :
Answer:
Explanation:
mass of rocket m = .5 kg
height of wall h = 40 m
initial horizontal velocity u = .5 m /s
horizontal acceleration = force / mass
= 20 / .5
a = 40 m /s²
Let rocket falls or covers 40 m vertically downwards in time t
h = 1/2 gt² , initial vertical velocity = 0
40 = 1/2 x 9.8 x t²
t = 2.8566 s
During this period it will cover horizontal distance with initial velocity of .5 m /s and acceleration a = 40m /s²
horizontal distance = ut + 1/2 at²
= .5 x 2.8566 + .5 x 40 x 2.8566²
= 1.4283 + 163.2033
= 164.63 m .
The distance from the base of the wall does the rocket land is 164.63 m .
Calculation of the distance:
A 500 g model rocket is resting horizontally at the top edge of a 40-m-high wall when it is accidentally bumped. The bump pushes it off the edge with a horizontal speed of 0.5 m/s
We know that
horizontal acceleration = force / mass
a = 20 / .5
a = 40 m /s²
Now
h = 1/2 gt² ,
Here initial vertical velocity = 0
So,
40 = 1/2 x 9.8 x t²
t = 2.8566 s
Now
horizontal distance = ut + 1/2 at²
= .5 x 2.8566 + .5 x 40 x 2.8566²
= 1.4283 + 163.2033
= 164.63 m .
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