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 + Answer

Arrangement Initial Speed (m/s) Time of Flight (s) 3.2 2.9 2.6 3.5 Short range, Horizontal Short range, 15° Short range, 15 Short range, 30° Short range, 45 0.23 0.29 0.42 0.57

Explanation / Answer

here,

launch height, h = 2.5 m

Part A :
horizonatal speed vx = 2.6 * Cos 30 = 2.25 m/s
vertical speed vy = 2.6 * Sin 30 = 1.3 m/s

Part B :
maximum time to reach max height, t1

t1 = vy/g
t1 = 1.3/9.8
t1 = 0.133 s

Part C:
maximum Height when launched from launched height, Hm
hm = (v^2*Sin^2A)/2g
hm = 2.6^2 * (Sin30)^2 / (2*9.8)
hm = 0.086 m

height from ground, hg = hm +h = 2.5 + 2.086 = 2.586 m

Part D:
time for falling maximum height, hm will be same as time to attain max height as motion is influnced by any other external force,
t2=t1 = 0.27 s

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