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Answer :
Final answer:
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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Final answer:
The projectile fired at an angle of 70 degrees with an initial speed of 39.4 m/s will hit a building at approximately 69.82 m.
Explanation:
The question pertains to the projectile motion of an object. To calculate the height at which the projectile strikes the building, we can use the equations of motion. First, we need to break down the initial velocity into its vertical component, which can be done using the relation: vy = v sin(θ), which yields 39.4 sin(70) ≈ 36.97 m/s.
Now, we can calculate the time taken to reach the highest point of the trajectory using the equation: t = vy/g, where g is the acceleration due to gravity (approximately 9.81 m/s2). This calculation yields approximately 3.77 seconds.
Considering that the projectile is fired from the ground level, the height hh is reached at the peak point of the trajectory. We can calculate this height using the formula: h = vy * t - 0.5 * g * t2. Plugging in the known values, we get: 36.97 m/s * 3.77 s - 0.5 * 9.81 m/s2 * (3.77 s)2 ≈ 69.82 m.
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