Answer :
In order to solve the problem, let's write the equations of motion first. Let's take the x-axis as the horizontal direction, and the y-axis on the vertical direction (pointing downward). Calling [tex]v_0=7.9 m/s[/tex] the initial velocity, and [tex]\alpha=18^{\circ}[/tex] the angle below the horizontal, the equations of motion are
[tex]S_x(t)=v_0 cos (\alpha )t[/tex]
[tex]S_y(t)=v_0 sin (\alpha ) t+ \frac{1}{2}gt^2 [/tex]
where [tex]g=9.81 m/s^2[/tex] is the gravitational acceleration.
(a) To find the distance covered by the ball horizontally, we must simply calculate Sx at the time the ball hits the ground (t=4.0 s):
[tex]S_x(4.0s)= 7.9 m/s \cdot cos (18^{\circ}) \cdot 4.0s=30.05 m[/tex]
(b) The height from which the ball was thrown is the value of Sy at 4.0s, which is the distance covered by the ball before hitting the ground:
[tex]S_y(4.0 s)=7.9 m/s \cdot sin(18^{\circ}) \cdot 4.0s + \frac{1}{2}(9.81 m/s^2)(4.0s)^2=88.24 m [/tex]
(c) To calculate how long does it take to the ball to reach 10.0 m below the initial point, we have to find the time at which Sy(t)=10.0 m. This means we must solve the following equation:
[tex]10.0m = v_0 sin (\alpha ) t + \frac{1}{2} gt^2 [/tex]
Using the data of the problem, we can solve this equation. We find two solutions for t: one is negative, so we can neglect it. The second one, which is the solution of the problem, is t=1.19 s.
[tex]S_x(t)=v_0 cos (\alpha )t[/tex]
[tex]S_y(t)=v_0 sin (\alpha ) t+ \frac{1}{2}gt^2 [/tex]
where [tex]g=9.81 m/s^2[/tex] is the gravitational acceleration.
(a) To find the distance covered by the ball horizontally, we must simply calculate Sx at the time the ball hits the ground (t=4.0 s):
[tex]S_x(4.0s)= 7.9 m/s \cdot cos (18^{\circ}) \cdot 4.0s=30.05 m[/tex]
(b) The height from which the ball was thrown is the value of Sy at 4.0s, which is the distance covered by the ball before hitting the ground:
[tex]S_y(4.0 s)=7.9 m/s \cdot sin(18^{\circ}) \cdot 4.0s + \frac{1}{2}(9.81 m/s^2)(4.0s)^2=88.24 m [/tex]
(c) To calculate how long does it take to the ball to reach 10.0 m below the initial point, we have to find the time at which Sy(t)=10.0 m. This means we must solve the following equation:
[tex]10.0m = v_0 sin (\alpha ) t + \frac{1}{2} gt^2 [/tex]
Using the data of the problem, we can solve this equation. We find two solutions for t: one is negative, so we can neglect it. The second one, which is the solution of the problem, is t=1.19 s.