![]() The height ℎ of the tower is, therefore, 28.224 metres. Then, we substitute these values into the equation of motion □ = □ □ + 1 2 □ □ : This means that our displacement and acceleration have the same sign. Same as the direction of vertical acceleration (namely, down). We can take the direction of vertical motion as being the Vertical acceleration is 9.8 m/s 2, and its time of flight isĢ.4 seconds. We know that the rock’s initial vertical velocity is 0, its Next, we want to calculate the vertical height of the tower. Thus, the distance □ between the base of the tower and the point where the rock landed is We substitute these values into the equation of motion We know that the rock’s initial horizontal velocity isĢ0.8 m/s, its horizontal acceleration is zero, and its time of flight Let us calculate the horizontal distance □ first. zero initial velocity and a constant acceleration of □ downward in the vertical direction.Notice that if we are given the time of a projectile’s flight, we can calculate both its horizontal and vertical displacements Giving us a horizontal distance of 12.95 m to 2 decimal places. Velocity of □ = 7 4 / m s and the horizontal acceleration of We now substitute this time into the equation of motion □ = □ □ + 1 2 □ □ using the initial horizontal Notice that the direction of acceleration is the same as the direction of motion, so the signs on □ andĠ. ![]() 8 / m s , and the vertical distance ofĠ.15 m into the equation of motion □ = □ □ + 1 2 □ □ . Thus, we substitute the initial vertical velocity of 0, the acceleration due to gravity of ![]() We first analyze the vertical motion of the arrow to find the total flight time of the arrow.
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