The figure below shows a vertical projectile. If the projectile's initial velocity (v1) is 200 m/s and the launch angle θ = 55°, determine the projectile's maximum vertical displacement (Δymax).
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The figure below shows a vertical projectile. If the projectile's initial velocity (v1) is 200 m/s and the launch angle θ = 55°, determine the projectile's maximum vertical displacement (Δymax).
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- A projectile's launch speed is 6.30 times its speed at maximum height. Find the launch angle θ0.How do you solve this? I've got 2/3 so farConsider a projectile launched at a height h feet above the ground and at an angle ? with the horizontal. If the initial velocity is v0 feet per second, the path of the projectile is modeled by the parametric equations x = t(v0 cos(?)) and y = h + (v0 sin ?)t - 16t2.Let h=6, v0=63.25, and ?=0.79. What is the range of the projectile (i.e., how far does it travel? Round your answers to two decimal places).
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