Question

What is the acceleration of a projectile at the top of its trajectory?

Answer 1

HEYA MATE HERE U GO.

❤. At the top of its projectory, the direction of acceleration is vertically downwards and its value is 9.8m^-2.

Answer 2

The acceleration at the top of a projectile is equal to g = 9.8 m/s².

Let's see the detailed explanation:

• Let's assume that a projectile is thrown with the initial velocity 'u' at an angle of $\theta$.

• The maximum height reached by the object in the projectile trajectory is $\boxed{H=\dfrac{u^{2}\sin^{2}(\theta)}{2g}}$

• At this max height (i.e. top of projectile), the Y axis velocity component will become zero and the X axis velocity will be $u\cos(\theta)$.

• The net acceleration of the object at that height will be equal to gravity (i.e. g = 9.8 m/s²), and the acceleration vector will be directed vertically downwards towards earth surface.

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