Answer:
Explanation:
Let the velocity of projectile be v and angle of throw be θ.
The projectile takes 5 s to touch the ground during which period it falls vertically by 100 m
considering its vertical displacement
h = - ut +1/2 g t²
100 = - vsinθ x 5 + .5 x 9.8 x 5²
5vsinθ =  222.5
vsinθ = 44.5
It covers 160 horizontally in 5 s
vcosθ x 5 = 160
v cosθ = 32
squaring and adding
v²sin²θ +v² cos²θ = 44.4² + 32²
v² = 1971.36 + 1024
v = 54.73 m /s
Answer:
55.42 m/s
Explanation:
Along the horizontal direction, the rock travels at constant speed: this means that its horizontal velocity is constant, and it is given by
u_x = d/t
Where
d = 160 m is the distance covered
t = 5.0 s is the time taken
Substituting, we get
u_x =160/5 = 32 m/s.
Along the vertical direction, the rock is in free-fall - so its motion is a uniform accelerated motion with constant acceleration g = -9.8 m/s^2 (downward). Therefore, the vertical distance covered is given by the
[tex]S=u_yt+\frac{1}{2}at^2[/tex]
where
S = -100 m is the vertical displacement
u_y is the initial vertical velocity
Replacing t = 5.0 s and solving the equation for u_y, we find
-100 = u_y(5) + (-9.81)(5)^2/2
u_y = 45.25 m/s
Therefore, the speed with which the rock was thrown u
[tex]u= \sqrt{u_x^2+u_y^2} \\=\sqrt{32^2+45.25^2}\\ = 55.42 m/s[/tex]
