Nathan is walking to the store and sees a snake slithering across the sidewalk. He jumps over it with an initial vertical velocity of 10 ft/s. How long does it take for Nathan to land back on the ground?
i need help really bad!
h = v0 t + 1/2 at^2
since acceleration from gravity is -10 m/s^2, when h=0, you have
10t - 5t^2 = 0
Now finish it off
To find the time it takes for Nathan to land back on the ground after jumping over the snake, we can use the equation for vertical motion:
s = ut + (1/2)gt^2
Where:
s is the displacement (height above the ground),
u is the initial vertical velocity,
g is the acceleration due to gravity (approximately -32 ft/s^2 for objects near the surface of the Earth),
t is the time.
Nathan's vertical displacement is zero when he lands back on the ground. His initial vertical velocity is 10 ft/s, and the acceleration due to gravity is -32 ft/s^2. Plugging in these values, we have:
0 = (10 ft/s)(t) + (1/2)(-32 ft/s^2)(t^2)
Simplifying the equation, we have:
0 = 10t - 16t^2
Rearranging the equation, we get:
16t^2 - 10t = 0
Factoring out t, we have:
t(16t - 10) = 0
Setting each factor equal to zero, we get:
t = 0 or t = 10/16 = 5/8
Since time cannot be zero for this scenario, the only valid solution is t = 5/8 seconds. Therefore, it takes Nathan 5/8 seconds to land back on the ground after jumping over the snake.
To find the time it takes for Nathan to land back on the ground, we need to consider the motion of his jump. We can use the kinematic equation for vertical displacement:
s = ut + (1/2) * a * t^2
where:
s = vertical displacement (0 ft, since Nathan lands back on the ground)
u = initial vertical velocity (10 ft/s, upwards in this case)
a = acceleration (acceleration due to gravity, which is approximately -32 ft/s^2, since it acts downwards)
t = time
Since Nathan jumps upwards and lands back on the ground, his final vertical velocity is zero.
At the highest point of his jump, the final velocity is zero and we can use the equation:
v = u + at
Rearranging this equation to solve for time (t):
t = (0 - u) / a
Substituting the given values:
t = (0 - 10 ft/s) / -32 ft/s^2
Simplifying the equation:
t = 10/32 s
Therefore, it takes approximately 0.3125 seconds (or about 0.31 seconds) for Nathan to land back on the ground.