Answer to Question #87815 in Mechanics | Relativity for Luis Garza

Question

Consider an object initially thrown at a 45 degree angle above horizontal. The object has an initial vertical velocity of 19.6 m/s. Neglecting air resistance, give the vertical and horizontal velocities and vertical and horizontal accelerations at each time 0 seconds, 1 second, 2 seconds, 3 seconds and 4 seconds

Answer 1

An object initially thrown at a 45 degree angle above horizontal, so its initial horizontal velocity is also 19.6 m/s. 

0 seconds:

$$v_y=19.6 \dfrac{m}{s}, v_x=19.6 \dfrac{m}{s}$$

$$a_x=0 \dfrac{m}{s^2}, a_y=-9.8 \dfrac{m}{s^2}$$

1 second:

$$v_y=19.6-9.8(1)=9.8 \dfrac{m}{s}, v_x=19.6 \dfrac{m}{s}$$

$$a_x=0 \dfrac{m}{s^2}, a_y=-9.8 \dfrac{m}{s^2}$$

2 seconds:

$$v_y=19.6-9.8(2)=0 \dfrac{m}{s}, v_x=19.6 \dfrac{m}{s}$$

$$a_x=0 \dfrac{m}{s^2}, a_y=-9.8 \dfrac{m}{s^2}$$

3 seconds:

$$v_y=19.6-9.8(3)=-9.8 \dfrac{m}{s}, v_x=19.6 \dfrac{m}{s}$$

$$a_x=0 \dfrac{m}{s^2}, a_y=-9.8 \dfrac{m}{s^2}$$

4 seconds:

$$v_y=19.6-9.8(4)=-19.6 \dfrac{m}{s}, v_x=19.6 \dfrac{m}{s}$$

$$a_x=0 \dfrac{m}{s^2}, a_y=-9.8 \dfrac{m}{s^2}$$