Question

Find the tangential and normal components of acceleration for a projectile fired at an angle \(\theta\) with the horizontal at an initial speed of \(v_{0}\). What are the components when the projectile is at its maximum height?

Short answer

The tangential acceleration of the projectile is \(0 ms^{-2}\), and the normal acceleration is \(g ms^{-2}\). At the maximum height, the direction of the normal acceleration is downward, but its magnitude remains as \(g ms^{-2}\)

Step-by-step solution

Identifying the Components of Acceleration

A projectile's acceleration at any point on its path can be broken down into two orthogonal components: the tangential component (\(a_t\)) and the normal component (\(a_n\)). The tangential acceleration results from changes in the magnitude of the velocity, while the normal acceleration results from changes in the velocity's direction. For a projectile, \(a_t\) is zero and \(a_n\) is equal to \(g\), the acceleration due to gravity, directed downward.

Calculating Components at Maximum Height

At the topmost point in its trajectory, a projectile's vertical speed is zero, and its velocity is entirely horizontal. Thus, at this point, the direction of motion (and thus of the tangential acceleration) is horizontal. Since for a projectile \(a_t\) is zero, it remains zero at the maximum height. The normal acceleration is perpendicular to the path and, at maximum height, it points downward. Therefore, at the maximum height, the normal acceleration (\(a_n\)) is still \(g\).

Final Conclusion

At any point in its trajectory, a projectile has a tangential acceleration of zero and a normal acceleration of \(g\). At the maximum height of its trajectory, the direction of the normal acceleration changes and becomes vertically downward, but its magnitude remains \(g\).