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
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[tex]m/s^{2}[/tex])
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
