Question

A football is thrown at an angle of 30.° above the horizontal. The magnitude of the horizontal

Answer 1

Answer:

1.53 s

Explanation:

Initially vertical component of velocity of the ball, uy = 7.5 m/s

Net displacement is vertical direction is zero, Δy =0

Use second equation of motion:

Δy = uy t + 0.5 a t²

Here, acceleration a = -g                           (g =9.8 m/s²)

Substitute all the values and solve for g

0 = 7.5 t -0.5 (9.8)t²

7.5 t = 4.9 t²

t = 1.53 s

Answer 2

Answer:

Explanation:

Let the initial velocity is u.

ux = 13 m/s  

uy = 7.5 m/s

Angle of projection, θ = 30°

ux = u Cos 30 = 13

u = 15 m/s

Height caught is same

use second equation of motion

h = uy t - 0.5 x gt²

0 = 75. t - 0.5 x 9.8 x t²

t = 1.24 s