# A projectile's launch speed is five times its speed at maximum height.Find launch angle.How do I find this?

Thus $u = 5u cos alpha implies cos alpha = 1/5#

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The launch angle ((\theta)) can be calculated using the formula:

[ \theta = \arctan\left(\frac{v^2}{gR}\right) ]

where: (v) = launch speed, (g) = acceleration due to gravity (approximately 9.8 m/s²), (R) = range of the projectile.

In this case, since the launch speed is five times the speed at maximum height, you can express (v) in terms of the speed at maximum height ((v_{\text{max}})):

[ v = 5 \cdot v_{\text{max}} ]

Substitute this expression for (v) into the launch angle formula to find the launch angle.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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