Question

You throw a baseball at an angle of 30.0 degrees above the horizontal. It reaches the highest point of its trajectory 1.05 s later. At what speed does the baseball leave your hand?

Answer 1

Answer:

21m/s

Explanation:

Using one of the equations of motion;

v = u + at

where;

v = final velocity of the baseball

u = initial velocity at which the baseball leaves your hand

a = acceleration due to gravity =  -g (since the baseball moves upwards)

t = time taken during the motion.

Also, since the object (baseball) moves upwards, the above equation can be resolved into it's vertical component as follows;

(v sin θ) = (u sin θ) + a t        ------------------------------(i)

where;

θ = angle of projection = 30.0⁰

But;

At the highest point of the trajectory (where the baseball reaches the maximum height), the velocity (v) is zero (0).

Therefore, from the question;

v = 0;

a = -g (which will be taken as 10$m/s^{2}$)

t = 1.05s

Substituting the above values of v, a and t into equation (i) gives;

=> 0 sin 30⁰ = u sin 30⁰ - (10 x 1.05)

=> 0 = 0.5u - 10.5

=> 0.5u = 10.5

=> u = 21m/s

Therefore, the speed at which the baseball leaves your hand is 21m/s

Answer 2

Answer:

The ball leaves your hand at 20.6 m/s

Explanation:

Hi there!

The equation of the vertical velocity of the ball is the following:

vy = v0 · sin α + g · t

Where:

vy = vertical component of the velocity vector at time t.

v0 = initial velocity.

α = throwing angle.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

t = time.

When the ball reaches the highest point of its trajectory, its vertical velocity is zero. So, using the equation f vertical velocity we can solve it for the initial velocity:

vy = v0 · sin α + g · t

0 m/s = v0 · sin(30°) - 9.8 m/s² · 1.05 s

9.8 m/s² · 1.05 s  / sin(30°) = v0

v0 = 20.6 m/s

The ball leaves your hand at 20.6 m/s