Problem check please (projectile motion)

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.

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HallsofIvy

Homework Helper
"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.)

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