The only good use of a fruitcake is in catapult practice. Curve 1 in Fig. 4-24 gives the height y of a catapulted fruitcake versus the angle between its velocity vector and its acceleration vector during flight. (a) Which of the lettered points on that curve corresponds to the landing of the fruitcake on the ground? (b)

Curve 2 is a similar plot for the same launch speed but for a different launch angle. Does the fruitcake now land farther away or closer to the launch point?

The angle between velocity and acceleration is $\theta$.

This problem is based on the concept of projectile motion. When a particle is thrown near the earth’s surface, it travels along a curved path under constant acceleration directed towards the center of the earth surface. Also it involves the range of the projectile which is the displacement in the horizontal direction Considering the angle between the velocity vector and acceleration and comparing it with the height at points given, it can be determined the landing point or launching point.

Formulae:

The range of the projectile is given by,

$R=\frac{v_0^2\times \sin \left(2\theta \right)}{g}$

We know the acceleration due to gravity is directed downward. From diagram we can say that, B is launching point because initially angle between velocity and acceleration is greater than $90°$and it will keep on getting smaller. And at maximum height angle between velocity and acceleration is $90°$As it starts coming down, the angle would further reduce and at the landing point, it would be smallest. The point of landing of the fruitcake on the ground is A.

From thecurve2, we can see that value of$\theta$at the launching is increased (Point B indicates the launching point as discussed above), it means the range will decrease.

Hence object would fall closer.