Answer:
1.3m/s²
Explanation:
Given values in the first part: acceleration a, time t₁:
1) velocity v₀ [tex] = at_1[/tex]
2) height h₀ [tex]=\frac{1}{2}at_1^2[/tex]
Given values in the second part: acceleration -g, time t₂:
3) height h [tex]= -\frac{1}{2}gt_2^2+v_0t_2+h_0[/tex]
Combining equations 1,2,3 and setting h to zero:
[tex]0=-\frac{1}{2}gt_2^2+(at_1)t_2+\frac{1}{2}at_1^2\\ 0=a(t_1t_2+\frac{1}{2}t_1^2)-\frac{1}{2}gt_2^2[/tex]
Solve for a with t₁ = 4s and t₂=2.1s:
[tex]a=\frac{1}{2}gt_2^2(\frac{1}{t_1t_2+\frac{1}{2}t_1^2})[/tex]
