A shotputter throws the shot with an initial speed of 16.0 m/s at a 36.0 angle to the horizontal. Calculate the horizontal distance traveled by the shot if it leaves the athlete's hand at a height of 2.00 m above the ground
Hmm, I hate physic's, it messes with your head.
Solve this equation for t.
Y = 2.00 + 16sin36*t - (g/2)t^2
and take the positive root for t.
g = 9.81 m/s^2, of course
Then multipy t by the horizontal velocity component, 16.0 cos 36 = 12.94 m/s
36.1 is the answer i got
its wrong though
what could i have done wrong?
To calculate the horizontal distance traveled by the shot, we can use the projectile motion equations. The horizontal distance traveled by the shot (range) can be calculated using the formula:
Range = (initial velocity * time) * cos(angle)
First, we need to find the time of flight, which is the total time the shot spends in the air. The formula for time of flight (t) is:
t = (2 * initial velocity * sin(angle)) / g
where:
g is the acceleration due to gravity (approximately 9.8 m/s^2).
Let's calculate the time of flight:
t = (2 * 16.0 m/s * sin(36.0°)) / 9.8 m/s^2
t = 2.92 s
Now, plug the time of flight into the range formula to calculate the horizontal distance traveled by the shot:
Range = (16.0 m/s * 2.92 s) * cos(36.0°)
Range = 45.21 m
Therefore, the horizontal distance traveled by the shot is approximately 45.21 meters.