Initial speed of baseball thrown at 35° from 2.0 m height landing in 3 s

[confused]

Hi all,

I just wanted to check up on a problem I just did.


The problem is:
A vall is thrown by a baseball player. The ball is released from the player's hand at an angle of 35 degrees above the horizontal direction at a height of 2.0 meters above the playing field. The ball lands on the field 3 seconds later after it leaves the players hand. What is the inital speed of the ball as it leaves the players hand

This is how I attempted to solve the problem, but not sure if it is correct or not.

known variables
y=2.0m
t=3.00s
a=-9.8m/s^2
angle = 35 degrees

first, find the inital speed in the y direction:
y=vo(t)+.5(g)(t)^2
2.0m=vo(3.00s)+.5(-9.8)(3.00s)^2
v=15.37 m/s

now make this a y-component vector and find the x component of this vector by taking the Vi of y and dividing it by (tan 35). I get 20.39 m/s. Now since the speed is the magnitude of the velocity, take the magnitude of both the x and y components of the Vi vector and I get:
25.35m/s.

Any help would be greatly appreciated.

 

[HallsofIvy]

"first, find the inital speed in the y direction:
y=vo(t)+.5(g)(t)^2
2.0m=vo(3.00s)+.5(-9.8)(3.00s)^2
v=15.37 m/s"

This is the initial vertical speed necessary for the ball to reach a height of 2.0 m ABOVE the ballplayers hand. If the ball was thrown from a height of 2.0 m above the ground, then it will hit the ground when v0(3 s)+ (1/2)(-9.8)3^2= -2.
(That's the same as taking the ground to be h=0, the initial height of the ball 2 m so you are solving 2+ v0(3)+ (1/2)(-9.8)3^2= 0.)

 

[M98Ranger]

Hi there,

Your method looks correct to me! You used the kinematic equation for displacement in the y-direction and solved for the initial velocity, which gave you a value of 15.37 m/s. Then, you used trigonometry to find the x-component of the initial velocity, which is 20.39 m/s. Finally, you used the Pythagorean theorem to find the magnitude of the initial velocity, which is 25.35 m/s.

One thing to keep in mind is the direction of the initial velocity. Since the ball is being thrown at an angle of 35 degrees above the horizontal direction, the initial velocity should have both an x and y component. So instead of just writing 25.35 m/s as the magnitude of the initial velocity, you could write it as (20.39 m/s, 15.37 m/s) to indicate the direction as well. Other than that, your solution looks good! Keep up the good work.