A projectile is shot vertically with initial speed v?, neglecting air resistance, and assuming that g is constant, find where the projectile lands when it hits the ground.
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The projectile is shot vertically, which means it goes straight up and then comes straight down. Show more…
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A projectile is fired at an angle θ with respect to ground. The projectile launch velocity is V m/s. Find the difference between projectile range when there is no air resistance and when air resistance is present. For the case when the air resistance is present assume that the projectile decelerates horizontally at A (m/s2). Express the difference in range between the two situations in terms of θ, A, V and g (where g is the downward acceleration due to gravity).
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A projectile is launched with an initial speed v0 at an angle Θ (here 0 ≤ Θ ≤ π/2). The height as a function of time, h(t), is shown in a graph below. The equation for h(t) is h(t) = -1/2gt^2 + (v0 sin Θ)t. Here g is a constant (the acceleration due to gravity). (a) Find the maximum height reached. (b) Find the time it takes to reach the ground again. [Hint: set h(t) = 0] (c) The distance it travels horizontally, x(t), is given by x(t) = (v0 cos Θ)t. How far has the projectile traveled (horizontally) by the time it reaches the ground again? Call this distance X. [Hint: use part (b)] (d) Notice the answer to (c) depends on the launch angle Θ, so we can think of X as a function of Θ. Find the angle that will result in the farthest distance traveled.
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A projectile is fired with initial speed $v_{0} \mathrm{~m} / \mathrm{s}$ from a height of $h$ meters at an angle of $\theta$ above the horizontal. Assuming that the only force acting on the object is gravity, find the maximum altitude, horizontal range and speed at impact. $$ v_{0}=49, h=0, \theta=\frac{\pi}{4} $$
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