Answer:
[tex]u=29.75\ m/s[/tex]
Explanation:
given,
ball is thrown at an angle = 45°
horizontal distance traveled by the ball = 296 ft (Range)
1 ft = 0.3048 m
296 ft = 296 x 0.3048 m
= 90.22 m
initial velocity of the ball = ?
using the formula of Range
[tex]R = \dfrac{u^2sin\ 2\theta}{g}[/tex]
[tex] 90.22 =\dfrac {u^2 sin\ 2(45^0)}{9.81}[/tex]
[tex]u^2(1) = 90.22 \times 9.81[/tex]
[tex]u^2 = 885.058[/tex]
[tex]u=\sqrt{885.058}[/tex]
[tex]u=29.75\ m/s[/tex]
hence, the initial velocity of the ball is [tex]u=29.75\ m/s[/tex]