Answer to Question #258926 in Physics for Jhon

Question

An object is launched at a velocity of 25 m/s in a direction making an angle of

30° upward with the horizontal.

a. What is the maximum height reached by the object?

b. What is the total flight time (between launch and touching the ground) of

the object?

c. What is the horizontal range (maximum x above the ground) of the object?

d. What is the magnitude of the velocity of the object just before it hits the

ground?

Answer 1

Given:

$v_0=25\:\rm m/s$

$g=9.8\:\rm m/s^2$

$\theta=30^{\circ}$

(a) The maximum height reached by the object

$$h_{\max}=\frac{v_0^2\sin^2\theta}{2g}=\frac{25^2\sin^230^{\circ}}{2*9.8}=8.0\:\rm m$$

(d) Time of flight

$$t=\frac{2v_0\sin\theta}{g}=\frac{2*25\sin30^{\circ}}{9.8}=2.6\:\rm s$$

(e) Range

$$h_{\max}=\frac{v_0^2\sin2\theta}{g}=\frac{25^2\sin60^{\circ}}{9.8}=48\:\rm m$$

(d) The final speed is equal to initial speed, so

$$v_f=25\:\rm m/s$$