Question

An object is thrown into the air with a velocity of 30.0 m/s at an angle of 35.0° to
the horizontal. What is the range of this object?

Answer 1

To find the range of the object, we first need to determine the time it takes for the object to reach its maximum height.

The vertical component of the initial velocity is given by:
v = v₀ * sin(θ)
v = 30 * sin(35°)
v ≈ 17.20 m/s

Using the equation v = u + at, where v = 0 (since the object momentarily stops at the maximum height) and a = -9.8 m/s² (acceleration due to gravity):
0 = 17.20 - 9.8t
9.8t = 17.20
t ≈ 1.76 s

The time it takes to reach maximum height is approximately 1.76 seconds. The total time in the air will be double this time, so that the object will fall back to the same height from which it was thrown.

Therefore, the total time in the air is 2 * 1.76 = 3.52 seconds.

Now we can find the horizontal distance traveled by the object using the equation:
range = horizontal velocity * time

The horizontal component of the initial velocity is given by:
u = v₀ * cos(θ)
u = 30 * cos(35°)
u ≈ 24.53 m/s

range = 24.53 m/s * 3.52 s
range ≈ 86.24 meters

Therefore, the range of the object is approximately 86.24 meters.