A man, located a distance [tex]d = 3.0 \, \text{m}[/tex] from a target, fires a projectile at an angle [tex]\theta = 70^\circ[/tex] above the horizontal. If the initial speed of the projectile is [tex]v = 39.4 \, \text{m/s}[/tex], at what height [tex]h[/tex] does the projectile strike the building?

The height at which the projectile would strike the target can be calculated using the formula for maximum height in a projectile motion, resulting in an approximate height of 70.76 meters above the ground.

Explanation:

In the field of Physics, this problem is a typical example of projectile motion. The height at which the projectile strikes the target can be found by using the vertical component of the motion. We break the initial speed into vertical and horizontal components using the launch angle θ. The vertical component of initial velocity (vy) can be calculated as v*sin(θ), which is 39.4*sin(70) = 37.18 m/s. The maximum height (h) attained by a projectile can be calculated by using the formula h = (vy²) / (2*g), where g is the acceleration due to gravity (approximately 9.8 m/s²).

So, inserting the values into the formula, h = (37.18*37.18) / (2*9.8), we get h ≈ 70.76 m. Therefore, the projectile would strike the target at a height of approximately 70.76 meters above the ground.

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