Answer:
-67.24 m/s²
Explanation:
From the given information:
Using the formula for Kinematic equation i,e [tex]v_y^2 -v_{0y}^2=2a_yd[/tex] to find the speed with which the pole-vaulter approaches the pad
where;
[tex]v_y =[/tex] the speed when the individual approaches the pad = ???
[tex]v_{oy}=[/tex] initial speed at the location of the bar = 0
[tex]a_y =[/tex] the individual acceleration which is also equivalent to acceleration due to gravity. = 9.8 m/s²
d = distance travelled
Thus;
d = 4.2m - 80 cm
d = 4.2 m - 0.8 m
d = 3.4 m
[tex]v_y^2 -(0)^2=2\times 9.8 \ m/s^2 \times 3.4 \ m[/tex]
[tex]v_y^2 =2\times 9.8 \ m/s^2 \times 3.4 \ m \\ \\ v_y ^2 = 66.64 \ m/s \\ \\ v_y = \sqrt{66.64 \ m/s} \\ \\ v_y = 8.2 \ m/s[/tex]
Thus, the initial speed of the pole-vaulter at the top of the pad = 8.2 m/s
Using the same above equation to determine the acceleration as the pole-vaulter approaches rest on the pad:
i.e
[tex]v_f^2 -v_{0}^2=2a_yd'[/tex]
[tex]0^2 -(-8.2\ m/s)^2=2a(0.5 \ m) \\ \\ a = - \dfrac{(8.2 \ m/s)^2}{2(0.5 \ m) } \\ \\ \mathbf{a = -67.24 \ m/s^2}[/tex]
