Will give brainliest! a shot put is thrown upward with a velocity of 35 ft./sec. at a height of 4 ft. and an angle of 40°. how long will it take for the shot put to be a horizontal distance of 40 ft. from the person throwing it? acceleration due to gravity is 32 ft./s^2 round your answer to the nearest hundreth

The answer is t = 1.49 s.

Explanation:

The motion of the shot put is a parabolic motion that can be divided into its component along the x- and the y-axis.

The question refers only to the horizontal distance (it doesn't say that the shot put must be on the ground), therefore, let's concentrate only on this component.

The horizontal component of the motion of a projectile (shot put) is a constant motion, given by the law:
x = x₀ + v₀ₓ · t

where:
v₀ₓ = v₀ · cosα

Therefore, in your case x₀ = 0 and the formula can be written as follows:
x = v₀ · cosα · t

Now, solve for t:
t = x / (v₀ · cosα)
  = 40 / (35 · cos40)
  = 1.492

Hence, the time taken to move horizontally 40 ft is 1.49 seconds.

Letter b sorry i’m i’m wrong xx