Answer:
yes
Explanation:
u = 30 m/s
θ = 45°
h = - 100 m (below)
d = 50 m
g = - 9.8 m/s^2
Use second equation of motion in vertical direction
[tex]h=u_{y}t +\frac{1}{2}a_{y}t^{2}[/tex]
[tex]-100=30\times Sin45\times t -0.5\times 9.8t^{2}[/tex]
[tex]-100 = 21.21 \times t -4.9 \times t^{2}[/tex]
[tex]t=\frac{21.21\pm \sqrt{21.21^{2}+4\times4.9 \times 100}}{9.8}[/tex]
By solving we get
t = 7.17 s
The horizontal distance traveled in this time
= u Cos45 x t = 30 x 0.707 x 7.17 = 152.1 m
This distance is more than the width of the river, So the ball crosses the river.
