Answer:
c. find the slope of the velocity time graph
Suvat
we have s, u, v and we want a
the suvat equation with these values in is: v^2 = u^2 - 2as
so a = (-v^2 + u^2)/-2s
plug numbers in
a = (-85^2 + 0^2)/-2*36 = 7225/72 = 100.3... ms^-2
The rider's horizontal motion, and how much ground he covers before he hits it, have nothing to do with how long he takes to hit the ground. The problem is simply: "How long does it take an object to fall 1.12 m from rest ?"
This seems like a good time to use this formula:
Distance fallen from rest = (1/2) (acceleration) (time)²
The problem doesn't tell us what planet the skateboarder is exercising on. I'm going to assume it's on Earth, where the acceleration of gravity is 9.8 m/s². And now, here's the solution to the problem I just invented:
1.12 m = (1/2) (9.8 m/s²) (time)²
Time² = (1.12 m) / (9.8 m/s²)
Time² = 0.1143 sec²
Time = √(0.1143 sec² )
<em>Time = 0.34 second</em>
Answer:
The height of the building is 88.63 m.
Explanation:
Given;
initial component of vertical velocity,
= 12 m/s sin 26° = 5.26 m/s
initial horizontal component of the velocity,
= 12 m/s cos 26° =10.786 m/s
horizontal distance traveled by the rock, x = 40.4 m
time of flight is calculated as;
x =
t
t = x / ![u_i](https://tex.z-dn.net/?f=u_i)
t = 40.4 / 10.786
t = 3.75 s
Determine the final vertical velocity of the ball;
![v_f = v_i + gt\\\\v_f = 5.26 + (9.8 *3.75)\\\\v_f = 42.01 \ m/s](https://tex.z-dn.net/?f=v_f%20%3D%20v_i%20%2B%20gt%5C%5C%5C%5Cv_f%20%3D%205.26%20%2B%20%289.8%20%2A3.75%29%5C%5C%5C%5Cv_f%20%3D%2042.01%20%5C%20m%2Fs)
Determine the height of the rock;
![v_f^2 = v_i^2 + 2gh\\\\h = \frac{v_f^2 - v_i^2}{2g}\\\\ h = \frac{(42.01)^2 - (5.26)^2}{2*9.8}\\\\h = 88.63 \ m](https://tex.z-dn.net/?f=v_f%5E2%20%3D%20v_i%5E2%20%2B%202gh%5C%5C%5C%5Ch%20%3D%20%5Cfrac%7Bv_f%5E2%20-%20v_i%5E2%7D%7B2g%7D%5C%5C%5C%5C%20h%20%3D%20%5Cfrac%7B%2842.01%29%5E2%20-%20%285.26%29%5E2%7D%7B2%2A9.8%7D%5C%5C%5C%5Ch%20%3D%2088.63%20%5C%20m)
Therefore, the height of the building is 88.63 m.