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------------------------------------------------ Calculate the modulus of the force acting on a 7.25 kg block thrown at 45° which hits the ground at a horizontal distance of 16 m. It is assumed that the hand has moved 1.5m and the block is touching the ground at the height where it left the hand answer = 429N

The modulus of the force acting on the block can be calculated using the following steps:

1. First, calculate the initial vertical velocity (v) of the block when it leaves the hand. We can use the formula v = sqrt(2gh), where g is the acceleration due to gravity (9.8 m/s^2) and h is the height at which the block leaves the hand (1.5 m). Therefore, v = sqrt(2 * 9.8 * 1.5) = 6.89 m/s.

2. Next, calculate the time of flight (t) of the block. We can use the formula t = 2v * sin(angle) / g, where angle is the launch angle (45°). Therefore, t = 2 * 6.89 * sin(45°) / 9.8 = 0.99 s.

3. Then, calculate the horizontal distance (d) traveled by the block. We can use the formula d = v * cos(angle) * t. Therefore, d = 6.89 * cos(45°) * 0.99 = 4.87 m.

4. Now, we can calculate the horizontal component of the initial velocity (vx) using the formula vx = v * cos(angle). Therefore, vx = 6.89 * cos(45°) = 4.87 m/s.

5. Finally, we can calculate the modulus of the force (F) acting on the block using the equation F = (m * (v^2)) / d, where m is the mass of the block (7.25 kg) and v is the horizontal component of the initial velocity (4.87 m/s). Therefore, F = (7.25 * (4.87^2)) / 4.87 = 429 N.

So, The force exerted on the block has a 429 N modulus.

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