Explanation:
It is given that,
Initial velocity of frog, u = 100 m/s
It jumps at an angle of 60 degrees.
(a) Let [tex]v_x\ and\ v_y[/tex] are the x and y components of velocity. It can be calculated as :
[tex]u_x=u\ cos\theta[/tex]
[tex]u_x=100\times \ cos(60)[/tex]
[tex]u_x=50\ m/s[/tex]
[tex]u_y=u\ sin\theta[/tex]
[tex]u_y=100\times \ sin(60)[/tex]
[tex]u_y=86.6\ m/s[/tex]
(b) Let y is the maximum height reached by the frog. It can be calculated using the third equation of motion as :
At maximum height, [tex]v_y=0[/tex]
[tex]v_y^2-u_y^2=2ay[/tex] and a = -g
[tex]-u_y^2=-2g[/tex]
[tex]y=\dfrac{u_y^2}{2g}[/tex]
[tex]y=\dfrac{86.6^2}{2\times 9.8}[/tex]
y = 382.63 meters
(c) Let t is the time taken by the frog to reach its maximum height. It can be calculated as :
[tex]v_y=u_y-gt[/tex]
[tex]0=u_y-gt[/tex]
[tex]t=\dfrac{u_y}{g}[/tex]
[tex]t=\dfrac{86.6\ m/s}{9.8\ m/s^2}[/tex]
t = 8.83 seconds
Hence, this is the required solution.