What is the direction and magnitude of its velocity 0.75s later and just before it lands.?

Question

a tower is 9.0m high and a bull's eye is a horizontal distance of 3.5 m from the launch point at the top of the tower. the pumpkin is thrown from the launch point and is given an initial horizontal speed of 3.3 m/s.

Answer 1

At time t after launch, let:

x be the horizontal distance covered,

y be the height above ground,

v1 be the horizontal velocity,

v2 be the vertical velocity (down),

V be the resultant velocity,

a be the angle between its direction and the horizontal.

v1 = 3.3 (constant)

v2 = 9.81t ...(1)

When t = 0.75s:

v2 = 7.3575 m/s.

V = sqrt(v1^2 + v2^2)

= 8.06 m/s.

a = arctan(7.3575 / 3.3)

= 65.8 deg.

x = 3.3t ...(2)

y = 9.0 - 9.81t^2 / 2

On landing, y = 0.

9.0 - 9.81t^2 / 2 = 0

t = 1.3545 s.

Substituting this value of t in (1) gives:

v2 = 13.288 m/s

V = sqrt(v1^2 + v2^2)

= 13.7 m/s.

a = arctan(v2 / v1)

= 76.1 deg.