A rock is thrown upward with a velocity of 19 meters per second from the top of a 21 meter high cliff and it misses the cliff on the way back down. When will the rock be 11 meters from the water below?
the height h at time t is
h(t) = 21 + 19t - 4.9t^2
so, plug in h=11 and solve for t:
11 = 21 + 19t - 4.9t^2
t ~= 4.35 sec
5.93
To solve this problem, we can use the kinematic equation for the displacement of an object undergoing free fall:
d = vit + 0.5at^2,
where:
- d is the displacement,
- vi is the initial velocity,
- t is the time,
- a is the acceleration due to gravity.
First, let's calculate the time it takes for the rock to reach its highest point. At the highest point, the velocity will be zero. Given that the initial velocity (vi) is 19 m/s and the acceleration due to gravity (a) is approximately -9.8 m/s² (negative because it acts downwards), we can use the following equation:
0 = 19 - 9.8t_highest.
Solving for t_highest:
9.8t_highest = 19,
t_highest = 19 / 9.8,
t_highest ≈ 1.94 seconds.
Now, let's find the time it takes for the rock to fall from the highest point to a height of 11 meters. Since we know the acceleration due to gravity is -9.8 m/s², we can use the equation:
11 = 0 + (0.5 * -9.8)t_fall^2.
Simplifying:
5t_fall^2 = 11,
t_fall^2 = 11 / 5,
t_fall = sqrt(11 / 5),
t_fall ≈ 1.32 seconds.
To find the total time when the rock is 11 meters from the water below, we add the time it takes to reach the highest point and the time it takes to fall from the highest point to a height of 11 meters:
Total time = t_highest + t_fall,
Total time ≈ 1.94 + 1.32,
Total time ≈ 3.26 seconds.
Therefore, the rock will be approximately 11 meters from the water below after 3.26 seconds.