CompuChip said:Note that when he hits the ground, h(s) = 0. So you are asked to find all values 0 < s < 5 for which h(s) = 0.
Now, assuming you meant h(s)= sin2(4s)+2sin(4s)+1, you can start by defining y = 4s and rewriting s to y.
To solve this question, you will need to use the equation h(t) = h0 + v0t - 1/2gt^2, where h0 is the initial height, v0 is the initial velocity, t is time, and g is the acceleration due to gravity. Plug in the given values and solve for t. This will give you the time at which Shantaro hits the ground. Repeat the process for each given value of t to calculate the number of times Shantaro hits the ground in 5 seconds.
The given values for this question are the initial height (h0), initial velocity (v0), and acceleration due to gravity (g).
Yes, you can use a calculator to solve this question. However, make sure to use the correct order of operations and input the values correctly to get an accurate answer.
Yes, you can use the formula h(t) = h0 + v0t - 1/2gt^2 to solve this question. This is a commonly used formula in trigonometry to calculate the height of an object at a given time.
Sure, let's say the initial height is 50 meters, the initial velocity is 10 meters per second, and the acceleration due to gravity is 9.8 meters per second squared. Plugging these values into the formula h(t) = 50 + 10t - 1/2(9.8)t^2, we get the equation h(t) = 50 + 10t - 4.9t^2. Solving for t, we get t = 0 and t = 10. This means that Shantaro hits the ground at 0 seconds and 10 seconds. Therefore, he hits the ground twice in 5 seconds.