The object of the long jump is to launch oneself as a projectile and attain the
ID: 2112249 • Letter: T
Question
The object of the long jump is to launch oneself as a projectile and attain the maximum horizontal range . Here we shall treat the long jumper as a particle even though the human body is fairly large compared to the size of the trajectory. Actually there is one point within the athlete's body, called the "center of mass", that behaves as a projected particle. Our analysis of projectile motion implies that the long jumper should try to maximize v0 and take off at an angle as close to 45.0 degrees as possible. However, it is easier to get a large value of vx0 (by a running start) than it is to get a large value of vy0 ; consequently theta is usually much less than 45.0 degrees. Suppose the jumper takes off with vx0 9.00 m/s and jumps with a value of vy0 suficient to reach a vertical height of 1.00 m. Find v0, '0, and the horizontal range. Things like Vx0 and theta
Explanation / Answer
H =Voy^2 / 2 g ========>Voy = 4.42m/s...............V0 = sqrt(Vx^2+Vy^2) = 9.24 m/s................... range = Vox*Voy/g= 4.05m............. theta = tan ^-1 (Vy/vx) = 26.1 deg