Answer:
Explanation:
Given
Height S = 8.4m
Initial velocity u = 24m/s
Required
Time it will take to reach the ground
Using the equation of motion
S = ut +1/2gt²
8.4 = 24t + 1/2(9.8)t²
8.4 = 24t+4.9t²
4.9t²+24t - 8.4 = 0
Multiply through by 10
49t²+240t-84 = 0
Factorize
t = -240±√240²-4(49(-84)/2(49)
t = -240±√57600+16464/98
t = -240±√74064/98
t = -240±272.15/98
t = 32.15/98
t = 0.328seconds
Hence it takes 0.328seconds before it lands on the ground below.
b) The horizontal distance is the range expressed as;
R = u²/g
R = 27²/9.8
R = 729/9.8.
R = 74.39m
a. The time it took before the motorcycle stunt driver lands on the ground below is 1.31 seconds.
b. The distance from the base of the cliff that the motorcycle stunt driver traveled while in the air is 31.44 meters.
Given the following data:
We know that the acceleration due to gravity (g) of an object on planet Earth is equal to 9.8 [tex]m/s^2[/tex]
a. To determine how long (time) it took before the motorcycle stunt driver lands on the ground below:
At maximum height, time is given by the formula:
[tex]Time = \sqrt{\frac{2H}{g} }[/tex]
Where:
Substituting the given parameters into the formula, we have;
[tex]Time = \sqrt{\frac{2\times 8.4}{9.8} }\\\\Time = \sqrt{\frac{16.8}{9.8} }\\\\Time =\sqrt{1.7413}[/tex]
Time = 1.31 seconds.
b. To determine how far (distance) away from the base of the cliff the motorcycle stunt driver traveled while in the air:
[tex]Horizontal\;distance = horizontal\;speed \times time\\\\Horizontal\;distance = 24 \times 1.31[/tex]
Horizontal distance = 31.44 meters.
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