Answer:
a. 2.0secs
b. 20.4m
c. 4.0secs
d. 141.2m
e. 40m/s, ∅= -30°
Explanation:
The following Data are giving
Initial speed U=40m/s
angle of elevation,∅=30°
a. the expression for the time to attain the maximum height is expressed as
[tex]t=\frac{usin\alpha }{g}[/tex]
where g is the acceleration due to gravity, and the value is 9.81m/s if we substitute values we arrive at
[tex]t=40sin30/9.81\\t=2.0secs[/tex]
b. the expression for the maximum height is expressed as
[tex]H=\frac{u^{2}sin^{2}\alpha }{2g} \\H=\frac{40^{2}0.25 }{2*9.81} \\H=20.4m[/tex]
c. The time to hit the ground is the total time of flight which is twice the time to reach the maximum height ,
Hence T=2t
T=2*2.0
T=4.0secs
d. The range of the projectile is expressed as
[tex]R=\frac{U^{2}sin2\alpha}{g}\\R=\frac{40^{2}sin60}{9.81}\\R=141.2m[/tex]
e. The landing speed is the same as the initial projected speed but in opposite direction
Hence the landing speed is 40m/s at angle of -30°