Based on the short range initial speed with a launch angle of 30°, calculate the

Question

Based on the short range initial speed with a launch angle of 30°, calculate the initial vertical velocity and the initial horizontal velocity.

Based on the initial vertical velocity, find the time t it takes for the projectile to reach its maximum height.

Calculate this maximum height based on the initial vertical velocity and the height of the launcher.

Find time it takes to fall from the max height by calculating twice the maximum height divided by the acceleration due to gravity, and then taking the square root.

Add time it takes to reach the maximum height to the time it takes to fall from the maximum height to find the total time in flight. Calculate the range based on the initial horizontal velocity with a launch angle of (30°), and the calculated total time in flight.

Place the time-of-flight pad at the horizontal distance you calculated for the range.

*********Please Provide Step by Step guide*********

Table 1 Height of Launcher, d Initial speed, short range, 30° 2.6 Initial Horizontal Speed Initial Vertical Speed Time to reach Max Height Maximum Height Time to Fall from Max Height Total Time in Flight Predicted Range 0.25 m m/s m/s m/s

Explanation / Answer

As v = u + at.

At maximum height vertical velocity has to be zero. Hence 0 = usin$ - gt. Hence t = usin$/g.

For total time, the time taken to reach maximum height and to reach back to the ground will be same. Hence total time of flight will be 2t. i.e 2usin$/g.

For horizontal range , it will continue moving horizontally as it is in air with constant velocity i.e for time of flight it will move with ucos$ in horizontal direction.

Hence range will be 2usin$×ucos$/g.

For 30°, range will be 3u2/2g.

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