Answer:
(a) 3.5 s
(b) 28.6 m/s
(c) 34.34 m/s
(d) 44.64 m/s
Explanation:
(a)
[tex]y=v_o t +0.5gt^{2}[/tex] where[tex] v_o[/tex] is initial velocity which is zero hence
[tex]y=0.5gt^{2}[/tex] where t is time and g is acceleration due to gravity taken as 9.81 m/s2
Making t the subject, [tex]t=\sqrt {\frac {2y}{g}}[/tex]
Substituting y for -60 m and g as -9.81 m/s2
[tex]t=\sqrt {\frac {2*-60}{-9.81}}= 3.497487 s[/tex] and rounding off
t=3.5 s
(b)
Let [tex]v_{h}[/tex] be horizontal component of velocity
Since the range [tex]R=t*v_{h}[/tex] then making [tex]v_{h}[/tex] the subject
[tex]v_{h}=\frac {R}{t}[/tex] and substituting R for 100m, t for 3.5 s then
[tex]v_{h}=\frac {100 m}{3.5 s}= 28.57143 m/s[/tex] and rounding off
[tex]v_{h}=28.6 m/s[/tex]
(c)
The vertical component of velocity before hitting ground
[tex]v_y=v_oy+ gt[/tex] but [tex]v_oy[/tex] is zero hence
[tex]v_y=gt[/tex] and substituting g for -9.81 m/s2 and t for 3.5 s
[tex]v_y=-9.81*3.5= -34.335[/tex] rounded off as -34.34 m/s
(d)
The velocity before it hits the ground will be
[tex]v=\sqrt {(28.6)^{2}+(34.34)^{2}}=44.64 m/s[/tex]
