Explanation:
Given that,
Initial velocity = 19.7 m/s
Angle = 45°
(a). We need to calculate the velocity at the top of the trajectory
Using formula of velocity at the top of the trajectory
[tex]v_{y}=u\cos\theta[/tex]
Put the value into the formula
[tex]v_{y}=19.7\cos45[/tex]
[tex]v_{y}=13.93\ m/s[/tex]
The top of the trajectory is 13.93 m/s.
(b). We need to calculate the maximum height
Using formula of height
[tex]h=\dfrac{(u\sin\theta)^2}{2g}[/tex]
Put the value into the formula
[tex]h=\dfrac{(19.7\sin45)^2}{2\times9.8}[/tex]
[tex]h=9.90\ m[/tex]
The maximum height is 9.90 m.
(c). We need to calculate the range
Using formula of range
[tex]R=\dfrac{u^2\sin2\theta}{g}[/tex]
Put the value into the formula
[tex]R=\dfrac{19.7^2\times\sin2\times45}{9.8}[/tex]
[tex]R=39.6\ m[/tex]
The range is 39.6 m.
(d). We need to calculate the time
Using formula of time
[tex]t=\dfrac{2u\sin\theta}{g}[/tex]
Put the value in to the formula
[tex]t=\dfrac{2\times19.7\sin45}{9.8}[/tex]
[tex]t=2.84\ sec[/tex]
The time is 2.84 sec.
Hence, This is the required solution.
