Alice and Tom dive from an overhang into the lake below. Tom simply drops straight down from the edge, but Alice takes a running start and jumps with an initial horizontal velocity of 25 m/s. Neither person experiences any significant air resistance. Compare the time it takes each of them to reach the lake below.

A) the speed of Alice is larger than that of Tom.
B) the splashdown speed of Alice is larger than that of Tom.
C) they will both have the same speed.
D) the speed of Tom will always be 9.8 m/s larger than that of Alice.
E) the speed of Alice will always be 25 m/s larger than that of Tom.

Respuesta :

The options are wrong. The correct options are:

A) Alice reaches the surface of the lake first.

B) Tom reaches the surface of the lake first.

C) Alice and Tom will reach the surface of the lake at the same time.

Answer:

C) Alice and Tom will reach the surface of the lake at the same time.

Explanation:

Given:

Initial velocity of Tom (u₁) = 0 m/s (Drops from the height)

Initial velocity of Alice (u₂) = 25 m/s (Horizontally)

The height is same for both of them.

The acceleration in the vertical direction is same for both of them and is equal to the acceleration due to gravity.

Now, consider the motion of Tom. The time taken can be obtained using the equation of motion. The equation of motion is:

[tex]h=u_1t_1+\frac{1}{2}gt_1^2\\\\h=0+\frac{1}{2}gt_1^2\\\\t_1=\sqrt{\frac{2h}{g}}-----(1)[/tex]

Now, consider the motion of Alice. The motion of Alice is a projectile motion with initial velocity acting in the horizontal direction only. The vertical component of initial velocity is 0 m/s.

Consider the motion in the vertical motion of Alice.

The initial vertical velocity [tex](u_{2y})=0\ m/s[/tex]

The vertical height is given as:

[tex]h=u_{2y}t_2^2+\frac{1}{2}gt_2^2\\\\h=0+\frac{1}{2}gt_2^2\\\\t_2=\sqrt{\frac{2h}{g}}----(2)[/tex]

As seen from equations (1) and (2), the time taken by Tom and Alice are same.

So, option (C) is correct.