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?

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)

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

f=gm1m2/d^2 it d

f is force g is gravitaiton constant, m1 is mass of first object, m2 is mass of second object. d is distance between two object