What is the projectile's launch speed?

[Marcelgluch]

I though i was starting to understand but now i don't know anymore.. so please help! i have tried to figure it out myself .. but i just don't know where to start

1.A bullet is fired from a rifle that is held 1.6m above the ground in a horizontal position. The initial speed of the bullet is 1100 m/s. Find (a) the time it takes the for the bullet to strike the ground and (b) the horizontal distance traveled by the bullet.

2.A baseball is hit into the air at an initial speed of 36.6m/s and an angle of 50.0 degrees about the horizontal. at the same time, the center fielder starts running away from the batter, and he catches the ball 0.914 m above the level at which it was hit. if the center fielder is initially 1.10x10^2 from the plate, what must be his average speed?\

3.The highest barrier that a projectile can clear is 13.5 m, when the projectile is launched at an angle of 15.0 degrees above the horizontal. What is the projectile's launch speed?

i would appreciate any help at all... even if its just a hint on how to get started on these questions! thanks!

 

[rock.freak667]

Read up on http://www.pa.uky.edu/~moshe/phy231/lecture_notes/projectile.html"

 

[tomcue]

For all three scenarios, we need to use the equations of motion for projectile motion, which are:

1. y = y0 + v0y*t - 1/2*g*t^2 (vertical motion)
2. x = x0 + v0x*t (horizontal motion)

Where:
y = vertical position
y0 = initial vertical position
v0y = initial vertical velocity
t = time
g = acceleration due to gravity (9.8m/s^2)
x = horizontal position
x0 = initial horizontal position
v0x = initial horizontal velocity

To find the projectile's launch speed, we need to know the initial velocity of the projectile (v0). This can be found using the following equation:
v0 = √(v0x^2 + v0y^2)

We can then use the given information to solve for v0 in each scenario.

1. To find the time it takes for the bullet to strike the ground, we can set y = 0 and solve for t:
0 = 1.6 + 0 - 1/2*9.8*t^2
t = √(2*1.6/9.8) = 0.4 seconds

To find the horizontal distance traveled by the bullet, we can use the given initial velocity (v0x = 1100m/s) and the time found above to solve for x:
x = 0 + 1100*0.4 = 440 meters

Therefore, the projectile's launch speed is 1100 m/s.

2. To find the center fielder's average speed, we need to use the given information to find the time it takes for the baseball to reach the center fielder's position. To do this, we can set y = 0.914m and solve for t:
0.914 = 0 + 36.6*sin(50)*t - 1/2*9.8*t^2
t = 3.97 seconds

We can then use this time to find the horizontal distance traveled by the baseball:
x = 0 + 36.6*cos(50)*3.97 = 100.6 meters

Since the center fielder starts at a distance of 110 meters from the plate, his average speed must be:
v = (110+100.6)/3.97 = 53.4 m