Question

3. A projectile leaves the ground with initial velocity of 30 m/s, at an angle of θ above the

horizontal. At maximum height, the velocity of the projectile is half of its initial velocity.

a) Calculate the angle, θ of the projectile motion [2]

b) Calculate the time, ttotal needed for the projectile to reach the ground. [3]

Answer 1

Solution.

$v_0=30 m/s;$

$v=0.5v_0;$

$\theta -?; t -?;$

$a) v_x=v_{0x}=v_0cos\theta\implies cos\theta=\dfrac{v_x}{v_{0x}}$ ;

$cos\theta=\dfrac{0.5\sdot30 m/s}{30 m/s}=0.5; \theta=60^o;$

$b) t=\dfrac{2v_0sin\theta}{g};$

$t=\dfrac{2\sdot30 m/s\sdot0.866}{9.8 m/s^2}=5.3s;$

Answer: $a)\theta=60^o;$

$b)t=5.3s.$