Question

Two projectiles are launched from the ground simultaneously with an initial velocity 5.00 m/s. The first projectile is sent off at an angle of 40° with respect to the horizontal, and the second projectile at an angle of 71°. Which projectile lands on the ground in a region between 1.9 m and 2.5 m from the launcher?

A. First Projectile
B. Second Projectile
C. Both Projectiles
D. Neither Projectile

Answer 1

The range of a projectile launched with an initial velocity of 5.00 m/s at an angle of 40° is more likely to fall within a region between 1.9 m and 2.5 m from the launcher than one launched at an angle of 71°, making the first projectile the correct choice.

Step-by-step explanation:

The question concerns the range of two projectiles launched with the same initial velocity but at different angles. The range of a projectile, which refers to the horizontal distance it travels before hitting the ground, is determined by both the speed and the angle of launch, given by the equation for projectile motion:

R = \( \frac{v^2 \sin(2\theta)}{g} \), where R is the range, v is the initial velocity, \( \theta \) is the launch angle, and g is the acceleration due to gravity.

Using this formula, with v = 5.00 m/s, g = 9.80 m/s2, and substituting the respective angles, we calculate the range for both projectiles. However, without doing the calculation, we can infer that the first projectile, launched at a lower angle (closer to the optimal 45° for maximum range), is more likely to fall within the specified region between 1.9 m and 2.5 m than the second projectile, launched at a steep angle of 71°.

Therefore, the answer is: A. First Projectile.